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**Introduction **

In earlier classes, you have learnt about congruence of two geometric figures, and also some basic theorems and results on the congruence of triangle. Two geometric figures having same shape and size are congruent to each other but two geometric figures having same shape are called similar. Two congruent geometric figures are always similar but the converse may or may not be true.

All regular polygons of same number of sides such as equilateral triangle, squares, etc. are similar. All circles are similar.

In some cases, we can easily notice that two geometric figures are not similar. For example, a triangle and a rectangle can never be similar. In case, we are given two triangles, they may appear to be similar but actually they may not be similar. So, we need some criteria to determine the similarity of two geometric figures. In particular, we shall discuss similar triangles.

**~ Historical Facts**

**Euclid** was a very great Greek mathematician born about 2400 years ago. He is called the father of geometry because he was the first to establish a school of mathematics in Alexandria. He wrote a book on geometry called "The Elements" which has 13 volumes and has been used as a text book for over 2000 years. This book was further systematized by the great mathematician of Greece like Thales, Pythagoras, Pluto and Aristotle.

Abraham Lincoln, as a young lawyer was of the view that this greek book was a splendid sharpner of human mind and improves his power of logic and language.

**A king once asked Euclid, "Isn't there an easier way to understand geometry" **

**Euclid replied : "There is no royal-road way to geometry. Every one has to think for himself when studying."**

**Thales (640-546 B.C.) **a Greek mathematician was the first who initiated and formulated the theoretical study of geometry to make astronomy a more exact science. He is said to have introduced geometry in Greece. He is believed to have found the heights of the pyramids in Egypt, using shadows and the principle of similar triangles. The use of similar triangles has made possible the measurements of heights and distances. He proved the well-known and very useful theorem credited after his name : **Thales Theorem. **

**~ Congruent Figures **

Two geometrical figures are said to be congruent, provided they must have same shape and same size.

Congruent figures are alike in every respect.

**Ex.** 1. Two squares of the same length.

2. Two circle of the same radii.

3. Two rectangles of the same dimensions.

4. Two wings of a fan.

5. Two equilateral triangles of same length.

**~ Similar Figures **

Two figures are said to be similar, if they have the same shape. Similar figures may differ in size. Thus, two congruent figures are always similar, but two similar figures need not be congruent.

**Ex.** 1. Any two line segments are similar.

2. Any two equilateral triangles are similar

3. Any two squares are similar.

4. Any two circles are similar.

We use the symbol '~' to indicate similarity of figures.

**~ Similar Triangles **

DABC and DDEF are said to be similar, if their corresponding angles are equal and the corresponding sides are proportional.

i.e., when ÐA = ÐD, ÐB = ÐE, ÐC = ÐF

and .

And, we write DABC ~ DDEF.

The sign '~' is read as 'is similar to'.

**Theorem-1 (Thales Theorem or Basic Proportionality Theorem) :** *If a line is drawn parallel to one side of a triangle intersecting the other two sides, then the other two sides are divided in the same ratio. *

**Given :** A DABC in which line l parallel to BC (DEBC) intersecting AB at D and AC at E.

**To prove :**

**Construction :** Join D to C and E to B. Through E drawn EF perpendicular to AB i.e., EF ^ AB and through D draw DG ^ AC.

**Proof :**

**STATEMENT REASON **

**1.** Area of (DADE) = (AD × EF) Area of D =

Area of (DBDE) = (BD × EF)

**2.** By 1.

**3.** Similarly

**4.** Area (DBDE) = Area (DCDE) Ds BDE and CDE are on the same base BC and between the same parallel lines DE and BC.

**5.** = By 3. & 4.

**6.** = By 1. & 5.

Hence proved.

**Theorem-2 (Converse of Basic Proportionality Theorem) :** *If a line divides any two sides of a triangle proportionally, the line is parallel to the third side. *

**Given :** A DABC and DE is a line meeting AB and AC at D and E respectively such that =

**To prove :** DEBC

**Proof :**

**STATEMENT REASON **

**1.** If possible, let DE be not parallel to BC.

Then, draw DFBC

**2.** = By Basic Proportionality Theorem.

**3.** = Given

**4.**= From 2 and 3.

Þ +1 = + 1 Adding 1 on both sides.

Þ By addition.

Þ AF + FC = AC and AE + EC = AC.

Þ FC = EC Þ E and F coincide.

But, DFBC. Hence DEBC.

Hence, proved.

**Ex.1** In the adjoining figure, DEBC.

(i) If AD = 3.4 cm, AB = 8.5 cm and AC = 13.5 cm, find AE.

(ii) If and AC = 9.6 cm, find AE.

**Sol.** (i) Since DEBC, we have

Þ = 5.4

Hence, AE = 5.4 cm.

(ii) Since DEBC, we have

Let AE = x cm. Then, EC = (AC _ AE) = (9.6 _ x) cm.

Þ 5x = 3(9.6 _ x)

Þ 5x = 28.8 _ 3x Þ 8x = 28.8 Þ x = 3.6.

AE = 3.6 cm.

**Ex.2** In the adjoining figure, AD = 5.6 cm, AB = 8.4 cm, AE = 3.8 cm and AC = 5.7 cm. Show that DEBC.

**Sol.** We have, AD = 5.6 cm, DB = (AB _ AD) = (8.4 _ 5.6) cm = 2.8 cm.

AE = 3.8 cm, EC = (AC _ AE) = (5.7 _ 3.8) cm = 1.9 cm.

and

Thus,

DE divides AB and AC proportionally.

Hence, DEBC

**Ex.3** In fig, and ÐPST = ÐPRQ. Prove that PQR is an isosceles triangle. ** [NCERT] **

**Sol.** It is given that

So, STQR [Theorem]

Therefore, ÐPST = ÐPQR [Corresponding angles] - (1)

Also, it is given that

ÐPST = ÐPRQ (2)

So, ÐPRQ = ÐPQR [From 1 and 2]

Therefore PQ = PR [Sides opposite the equal angles]

i.e., PQR is an isosceles triangle.

**Ex.4** Prove that any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally (i.e., in the same ratio).

or

ABCD is a trapezium with DCAB. E and F are points on AD and BC respectively such that EFAB. Show that ** [NCERT] **

**Sol.** We are given trapezium ABCD.

CDBA

EFAB and CD both

We join AC.

It mets EF at O.

In DACD, OECD

Þ = ...(i)

(Basic Proportionality Theorem)

In DCAB, OFAB

Þ [B.P.T]

Þ ...(ii)

From (i) and (ii)

Hence, proved.

**Ex.5** Prove that the internal bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.

**(Internal Angle Bisector Theorem) **

**Sol.** **Given : ** A DABC in which AD is the internal bisector of ÐA.

**To Prove : **

**Construction :** Draw CEDA, meeting BA produced at E.

**Proof :**

**STATEMENT REASON **

**1.** Ð1 = Ð2 AD is the bisector of ÐA

**2.** Ð2 = Ð3 Alt. Ðs are equal, as CEDA and AC is the transversal

**3.** Ð1 = Ð4 Corres. Ðs are equal, as CEDA and BE is the transversal

**4.** Ð3 = Ð4 From 1, 2 and 3.

**5.** AE = AC Sides opposite to equal angles are equal

**6. **In DBCE, DACE

Þ = By B.P.T.

Þ = Using 5

Hence, Proved.

**Remark :** The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle. i.e., if in a DABC, AD is the bisector of the exterior of angle ÐA and intersect BC produced in D, .

**~ Axioms of similarity of triangles **

**1.** ** AA (Angle-Angle) Axiom of Similarity : **

If two triangles have two pairs of corresponding angles equal, then the triangles are similar. In the given figure, DABC and DDEF are such that

ÐA = ÐD and ÐB = ÐE.

DABC ~ DDEF

**2.** ** SAS (Side-Angle-Side) Axiom of Similarity : **

If two triangles have a pair of corresponding angles equal and the sides including them proportional, then the triangles are similar.

In the given fig, DABC and DDEF are such that

ÐA = ÐD and

DABC ~ DDEF.

**3.** ** SSS (Side-Side-Side) Axiom of Similarity : **

If two triangles have three pairs of corresponding sides proportional, then the triangles are similar.

If in DABC and DDEF we have :

, then DABC ~ DDEF.

**Ex.6** In figure, find ÐL.

**Sol.** In DABC and DLMN,

and

Þ

Þ DABC ~ DLMN (SSS Similarity)

Þ ÐL = ÐA = 180° _ ÐB _ ÐC

= 180° _ 50° _ 70° = 60°

ÐL = 60°

**Ex.7** In the figure, AB ^ BC, DE ^ AC, and GF ^ BC. Prove that DADE ~ DGCF.

**Sol.** Ð1 + Ð4 = Ð1 + Ð2 (each side = 90°)

Þ Ð4 = Ð2

Þ ÐA = ÐG ...(i)

Also ÐE = ÐF ...(ii) (each equal to 90°)

From (i) and (ii), we get AA similarity for triangles ADE and GCF.

Þ DADE ~ DGCF

**Ex.8** In fig, and Ð1 = Ð2. Prove that DPQS ~ DTQR.

**Sol.** Ð1 = Ð2 (Given)

Þ PR = PQ ...(i)

(Sides opposite to equal angles in DQRP)

Also (Given) ...(ii)

Form (i) and (ii), we have

Þ ...(iii)

Now, in triangles PQR and TQR, we have

ÐPQS = ÐTQR (each = Ð1)

and (from (3))

Þ DPQS ~ DTQR (SAS Similarity)

**Ex.9** In fig, CD and GH are respectively, the medians of DABC and DFEG, If DABC ~ DFEG, prove that

(i) DADC ~ DFHG

(ii) **(NCERT) **

**Sol.** DABC ~ DFEG (given)

Þ ÐA = ÐF, ...(i) (Q the corresponding angles of the similar triangles are equal)

Also, (Corresponding sides are proportional)

Þ

Þ

Þ ...(ii)

Now, in triangles ADC and FHG, we have

ÐA = ÐF and (By (i) and (ii))

Þ DADC ~ DFHG ** **(SAS similarity)

(ii) DADC ~ DFHG

Þ (Corresponding sides proportional)

Þ Þ

**Ex.10** ABC is a right triangle, right angled at B. If BD is the length of the perpendicular drawn from B to AC. Prove that:

(i) DADB ~ DABC and hence AB^{2} = AD × AC (ii) DBDC ~ DABC and hence BC^{2} = CD × AC

(iii) DADB ~ DBDC and hence BD^{2} = AD × DC (iv)

**Sol.** **Given :** ABC is right angled triangle at B and BD ^ AC

**To prove : **

(i) DADB ~ DABC and hence AB^{2} = AD × AC

(ii) DBDC ~ DABC and hence BC^{2} = CD × AC

(iii) DADB ~ DBDC and hence BD^{2} = AD × DC

(iv)

**Proof :** (i) In two triangles ADB and ABC, we have :

ÐBAD = ÐBAC (Common)

ÐADB = ÐABC (Each is right angle)

ÐABD = ÐACB (Third angle)

ÐADB ~ ÐABC (AAA Similarity)

Triangle ADB and ABC are similar and so their corresponding sides must be proportional.

Þ Þ AB × AB = AC × AD Þ AB^{2} = AD × AC This proves (a).

(ii) Again consider two triangles BDC and ABC, we have

ÐBCD = ÐACB (Common)

ÐBDC = ÐABC (Each is right angle)

ÐDBC = ÐBAC (Third angle)

Triangle are similar and their corresponding sides must be proportional.

i.e., DBDC ~ DABC

Þ Þ BC × BC = DC × AC Þ BC^{2} = CD × AC This proves (ii)

(iii) In two triangles ADB and BDC, we have :

ÐBDA = ÐBDC = 90°

Ð3 = Ð2 = 90° _ Ð1 [Q Ð1 + Ð2 = 90°, Ð1 + Ð3 = 90°]

Ð1 = Ð4 = 90° _ Ð2 [Q Ð1 + Ð2 = 90°, Ð2 + Ð4 = 90°]

DADB ~ DBDC (AAA criterion of similarity)

Þ Their corresponding sides must be proportional.

Þ Þ BD × BD = AD × DC

BD^{2} = AD × DC

Þ BD is the mean proportional of AD and DC

(iv) From (i), we have : AB^{2} = AD × AC

(ii), we have : BC^{2} = CD × AC

(iii), we have : BD^{2} = AD × DC

Consider

= (from (iii))

Thus we have proved the following :

*If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then: *

*(a) The triangle on each side of the perpendicular are similar to each other and also similar to the original triangle. *

*i.e., DADB ~ DBDC, DADB ~ DABC, DBDC ~ DABC *

*(b) The square of the perpendicular is equal to the product of the length of two parts into which the hypotenuse is divided by the perpendicular i.e., BD ^{2} = AD × DC.*

**~ RESULTS ON AREA OF SIMILAR TRIANGLES **

**Theorem-3 : the areas of two similar triangles are proportional to the squares on their corresponding sides. **

**Given :** DABC ~ DDEF

**To prove :**

**Construction :** Draw AL ^ BC and DM ^ EF.

**Proof :**

**STATEMENT REASON **

**1.** Area of D = × Base × Height

Þ

**2.** In DALB and DDME, we have

(i) ÐALB = ÐDME Each equal to 90°

(ii) ÐABL = ÐDEM DABC ~ DDEF Þ ÐB = ÐE

DALB ~ DDME AA-axiom

Þ Corresponding sides of similar Ds are proportional.

**3.** DABC ~ DDEF Given.

Þ Corresponding sides of similar Ds are proportional.

**4.** From 2 and 3.

**5.** Substituting in 1, we get :

**6.** Combining 3 and 5, we get :

**Corollary-1 : the areas of two similar triangles are proportional to the squares on their corresponding altitude. **

**Given :** DABC ~ DDEF, AL ^ BC and DM ^ EF.

**To prove :**

**Proof :**

**STATEMENT REASON **

**1.** Area of D = × Base × Height

Þ

**2.** In DALB and DDME, we have

(i) ÐALB = ÐDME Each equal to 90°

(ii) ÐABL = ÐDEM DABC ~ DDEF Þ ÐB = ÐE

Þ DALB ~ DDME AA-axiom

Þ Corresponding sides of similar Ds are proportional.

**3.** DABC ~ DDEF Given.

Þ Corresponding sides of similar Ds are proportional.

**4.** From 2 and 3.

**5. **Substituting in 1, we get :

.

Hence, proved.

**Corollary-2 : the areas of two similar triangles are proportional to the squares on their corresponding medians. **

**Given :** DABC ~ DDEF and AP, DQ are their medians.

**To prove :**

**Proof :**

**STATEMENT REASON **

**1.** DABC ~ DDEF Given

Þ .......I. Areas of two similar Ds are proportional to the

squares on their corresponding sides.

**2.** DABC ~ DDEF

Þ ......II Corresponding sides of similar Ds are proportional.

**3.** and ÐA = ÐD From II and the fact the DABC ~ DDEF

Þ DAPB ~ DDQE By SAS-similarity axiom

Þ .......III

Þ From II and III.

Þ ..........IV

**4.** From I and IV.

Hence, proved.

**Corollary-3 : The areas of two similar triangles are proportional to the squares on their corresponding angle bisector segments. **

**Given :** DABC ~ DDEF and AX, DY are their

bisectors of ÐA and ÐD respectively.

**To prove :**

**Proof :**

**STATEMENT REASON **

**1.** Areas of two similar Ds are proportional to the squares of the corresponding sides.

**2.** DABC ~ DDEF Given

Þ ÐA = ÐD

Þ ÐA = ÐD

Þ ÐBAX = ÐEDY ÐBAX = ÐA and ÐEDY = ÐD

**3.** In DABX and DDEY, we have Given

ÐBAX = ÐEDY From 2.

ÐB = ÐE DABC ~ DDEF

DABX ~ DDEY By AA similarity axiom

Þ Þ

**4.** From 1 and 3.

Hence, proved.

**Ex.11** It is given that DABC ~ DPQR, area (DABC) = 36 cm^{2} and area (DPQR) = 25 cm^{2}. If QR = 6 cm, find the length of BC.

**Sol.** We know that the areas of similar triangles are proportional to the squares of their corresponding sides.

Let BC = x cm. Then,

Û Û x^{2} = Û x = = = 7.2.

Hence BC = 7.2 cm

**Ex.12** P and Q are points on the sides AB and AC respectively of DABC such that PQBC and divides DABC into two parts, equal in area. Find PB : AB.

** **

**Sol.** Area (DAPQ) = Area (trap. PBCQ) [Given]

Þ Area (DAPQ) = [Area (DABC) _ Area (DAPQ)]

Þ 2 Area (DAPQ) = Area (DABC)

Þ ...(i)

Now, in DAPQ and DABC, we have

ÐPAQ = ÐBAC [Common ÐA]

ÐAPQ = ÐABC [PQBC, corresponding Ðs are equal]

DAPQ ~ DABC.

We known that the areas of similar Ds are proportional to the squares of their corresponding sides.

Þ [Using (i)]

Þ i.e., AB = · AP

Þ AB = (AB _ PB) Þ PB = ( _ 1) AB

Þ .

PB : AB = ( _ 1) :

**Ex.13 **Two isosceles triangles have equal vertical angles and their areas are in the ratio 16 : 25. Find the ratio of their corresponding heights.

**Sol.** Let DABC and DDEF be the given triangles in which AB = AC, DE = DF, ÐA = ÐD and

Draw AL ^ BC and DM ^ EF

Now, = 1 and = 1 [Q AB = AC and DE = DF]

Þ .

In DABC and DDEF, we have

and ÐA = ÐD

Þ DABC ~ DDEF [By SAS similarity axiom]

But, the ratio of the areas of two similar Ds is the same as the ratio of the squares of their corresponding heights.

Þ Þ

AL : DM = 4 : 5, i.e., the ratio of their corresponding heights = 4 : 5.

**Ex.14** If the areas of two similar triangles are equal, prove that they are congruent.

**Sol.** Let DABC ~ DDEF and area (DABC) = area (DDEF).

Since the ratio of the areas of two similar Ds is equal to the ratio of the squares on their corresponding sides, we have

=

Þ = 1 [Q Area (DABC) = Area (DDEF)]

Þ AB^{2} = DE^{2}, AC^{2} = DF^{2} and BC^{2} = EF^{2}

Þ AB = DE, AC = DF and BC = EF

DABC DDEF [By SSS congruence]

**Ex.15** In fig, the line segment XY is parallel to side AC of DABC and it divides the triangle into two parts of equal areas. Find the ratio . ** [NCERT] **

**Sol.** We are given that XYAC.

Þ Ð1 = Ð3 and Ð2 = Ð4 [Corresponding angles]

Þ DBXY ~ DBAC [AA similarity]

Þ [By theorem] ...(i)

Also, we are given that

ar (DBXY) = × ar (DBAC) Þ ...(ii)

From (i) and (ii), we have Þ = ...(iii)

Now, = 1 _ = 1 _ = 1 _ [By (iii)]

=

Hence,

**Theorem-4 [Pythagoras Theorem] : In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides. **

**Given : **A DABC in which ÐB = 90°.

**To prove :** AC^{2} = AB^{2} + BC^{2}.

**Construction : **From B, Draw BD ^ AC.

**Proof :**

**STATEMENT REASON **

**1. **In DADB and DABC, we have :

ÐBAD = ÐCAB = ÐA Common

ÐADB = ÐABC Each = 90°

DADB ~ DABC By AA axiom of similarity

Þ Corr. sides of similar Ds are proportional

Þ AB^{2} = AD × AC ..(i)

**2.** In DCDB and DCBA, we have :ÐCDB = ÐCBA Each = 90°

ÐBCD = ÐACB = ÐC Common

DCDB ~ DCBA By AA axiom of similarity

Þ Corr. sides of similar Ds are proportional

Þ BC^{2} = DC × AC ..(ii)

**3.** Adding (i) and (ii), we get

AB^{2} + BC^{2} = AD × AC + DC × AC

= (AD + DC) × AC = AC^{2 }Q AD + DC = AC

Hence, AB^{2} + BC^{2} = AC^{2}.

**Theorem-5 [ Converse of pythagoras Theorem] : In a triangle if the square of one side is equal to the sum of the squares of the other two sides, then the triangle is right angled. **

**Given : **A DABC in which AB^{2} + BC^{2} = AC^{2}

**To prove :** ÐB = 90°

**Construction : **Draw a DDEF in which

DE = AB, EF = BC and ÐE = 90°

**Proof :**

**STATEMENT REASON **

**1. **In DDEF, we have: ÐE = 90°

DE^{2} + EF^{2} = DF^{2} By Pythagoras Theorem

Þ AB^{2} + BC^{2} = DF^{2} Q DE = AB and EF = BC

Þ AC^{2} = DF^{2} Q AB^{2} + BC^{2} = AC^{2} (Given)

Þ AC = DF

**2.** In DABC and DDEF, we have :

AB = DE By construction

BC = EF By construction

AC = DF Proved above

DABC DDEF By SSS congruence

Þ ÐB = ÐE c.p.c.t

Þ ÐE = 90° Q ÐE = 90°

Hence, ÐB = 90°

**Ex.16** If ABC is an equilateral triangle of side a, prove that its altitude =

**Sol.** DABC is an equilateral triangle.

We are given that AB = BC = CA = a. AD is the altitude, i.e., AD ^ BC.

Now, in right angled triangles ABD and ACD, we have

AB = AC [Given]

and AD = AD [Common side]

Þ DABD DACD [By RHS congruence]

Þ BD = CD Þ BD = DC = BC =

From right triangle ABD,

AB^{2} = AD^{2} + BD^{2} Þ a^{2 }= AD^{2} +

Þ AD^{2} = a^{2} _

Þ AD = a

**Ex.17** In a DABC, obtuse angled at B, if AD is perpendicular to CB produced, prove that :

AC^{2} = AB^{2} + BC^{2} + 2BC × BD

**Sol.** In DADB, ÐD = 90°.

AD^{2} + DB^{2} = AB^{2} ... (i) [By Pythagoras Theorem]

In DADC, ÐD = 90°

AC^{2} = AD^{2} + DC^{2} [By Pythagoras Theorem]

= AD^{2} + (DB + BC)^{2}

= AD^{2} + DB^{2} + BC^{2} + 2DB × BC

= AB^{2} + BC^{2} + 2BC × BD [Using (i)]

Hence, AC^{2} = AB^{2} + BC^{2} + 2BC × BD.

**Ex.18** In the given figure, ÐB = 90°. D and E are any points on AB and BC respectively. Prove that :

AE^{2} + CD^{2} = AC^{2} + DE^{2}. ** [NCERT] **

**Sol.** In DABE, ÐB = 90°

AE^{2} = AB^{2} + BE^{2} ...(i)

In DDBC, ÐB = 90°.

CD^{2} = BD^{2} + BC^{2} ...(ii)

Adding (i) and (ii), we get :

AE^{2} + CD^{2} = (AB^{2} + BC^{2}) + (BE^{2} + BD^{2})

= AC^{2} + DE^{2} [By Pythagoras Theorem]

Hence, AE^{2} + CD^{2} = AC^{2} + DE^{2}.

**Ex.19** A point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D.

Prove that: OA^{2} + OC^{2} = OB^{2} + OD^{2}.

**Sol.** Through O, draw EOFAB. Then, ABFE is a rectangle.

In right triangles OEA and OFC, we have:

OA^{2} = OE^{2} + AE^{2}

OC^{2} = OF^{2} + CF^{2}

OA^{2} + OC^{2} = OE^{2} + OF^{2} + AE^{2} + CF^{2} ... (i)

Again, in right triangles OFB and OED, we have :

OB^{2} = OF^{2} + BF^{2}

OD^{2} = OE^{2} + DE^{2}

OB^{2} + OD^{2} = OF^{2}+ OE^{2} + BF^{2} + DE^{2}

= OE^{2} + OF^{2} + AE^{2} + CF^{2} ...(ii) [Q BF = AE & DE = CF]

From (i) and (ii), we get

OA^{2} + OC^{2} = OB^{2} + OD^{2}.

**Ex20 **In the given figure, DABC is right-angled at C.

Let BC = a, CA = b, AB = c and CD = p, where CD ^ AB.

Prove that: (i) cp = ab (ii) ** **

**Sol.** (i) Area of DABC = AB × CD = cp.

Also, area of DABC = BC × AC = ab.

cp = ab. Þ cp = ab

(ii) cp = ab Þ p =

Þ p^{2} =

Þ [Q c^{2} = a^{2} + b^{2}]

Þ .

**Ex.21** Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.** (Appollonius Theorem) **

**Sol.** Given: A DABC in which AD is a median.

To prove : AB^{2} + AC^{2} = 2AD^{2} + 2 or AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

Construction : Draw AE ^ BC.

Proof : Q AD is median

BD = DC

Now, AB^{2} + AC^{2} = (AE^{2} + BE^{2}) + (AE^{2} + CE^{2}) = 2AE^{2} + BE^{2} + CE^{2 }

= 2[AD^{2} _ DE^{2}] + BE^{2} + CE^{2 }

= 2AD^{2} _ 2DE^{2} + (BD + DE)^{2} + (DC _ DE)^{2 }

= 2AD^{2} _ 2DE^{2} + (BD + DE)^{2} + (BD _ DE)^{2 }

= 2(AD^{2} + BD^{2}) = = 2AD^{2} + 2

Hence, Proved.

**~ Synopsis **

**8 SIMILAR TRIANGLES**. Two triangles are said to be similar if

(i) Their corresponding angles are equal and (ii) Their corresponding sides are proportional.

**8 **All congruent triangles are similar but the similar triangles need not be congruent.

**8 **Two polygons of the same numbers of sides are similar, if

(i) their corresponding angles are equal and

(ii) their corresponding sides are in the same ratio.

**8 BASIC PROPORTIONALITY THEOREM.** In a triangle, a line drawn parallel to one side, to intersect the other sides in distinct points, divides the two sides in the same ratio.

**8 CONVERSE OF BASIC PROPORTIONALITY THEOREM.** If a line divides any two sides of a triangle in the same ratio, the line must be parallel to the third side.

**8 AAA-SIMILARITY.** If in two triangles, corresponding angles are equal, i.e., the two corresponding angles are equal, then the triangles are similar.

**8 SSS-SIMILARITY.** If the corresporiding sides of two triangles are proportional, then they are similar.

**8 SAS-SIMILARITY.** If in two triangles one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

**8** The ratio of the areas of similar triangles is equal to the ratio of the squares of their corresponding sides.

**8 PYTHAGORAS THEOREM.** In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

**8 CONVERSE OF PYTHAGORAS THEOREM.** In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.

**Exercise-1 (for school/board exams) **

**Objective type Questions **

**Choose The Correct One **

**1.** Triangle ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. Triangle DEF is similar to DABC. If EF = 4 cm, then the perimeter of DDEF is :

(A) 7.5 cm (B) 15 cm (C) 22.5 cm (D) 30 cm

**2.** In DABC, AB = 3 cm, AC = 4 cm and AD is the bisector of ÐA. Then, BD : DC is :

(A) 9 : 16 (B) 16 : 9 (C) 3 : 4 (D) 4 : 3

**3.** In an equilateral triangle ABC, if AD ^ BC, then:

(A) 2AB^{2} = 3AD^{2} (B) 4AB^{2} = 3AD^{2} (C) 3AB^{2} = 4AD^{2} (D) 3AB^{2} = 2AD^{2}

**4.** ABC is a triangle and DE is drawn parallel to BC cutting the other sides at D and E. If AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm, then AE is equal to :

(A) 1.4 cm (B) 1.8 cm (C) 1.2 cm (D) 1.05 cm

**5.** The line segments joining the mid points of the sides of a triangle form four triangles each of which is :

(A) similar to the original triangle. (B) congruent to the original triangle.

(C) an equilateral triangle. (D) an isosceles triangle.

**6.** In DABC and DDEF , ÐA = 50°, ÐB = 70°, ÐC = 60°, ÐD = 60°, ÐE = 70°, ÐF = 50°, then DABC is similar to:

(A) DDEF (B) DEDF (C) DDFE (D) DFED

**7.** D, E, F are the mid points of the sides BC, CA and AB respectively of DABC. Then DDEF is congruent to triangle

(A) ABC (B) AEF (C) BFD, CDE (D) AFE, BFD, CDE

**8.** If in the triangles ABC and DEF, angle A is equal to angle E, both are equal to 40°, AB : ED = AC : EF and angle F is 65°, then angle B is :-

(A) 35° (B) 65° (C) 75° (D) 85°

**9.** In a right angled DABC, right angled at A, if AD ^ BC such that AD = p, If BC = a, CA = b and AB = c, then:

(A) p^{2} = b^{2} + c^{2} (B)

(C) (D) p^{2} = b^{2 }c^{2}

**10.** In the adjoining figure, XY is parallel to AC. If XY divides the triangle into equal parts, then the value of =

(A) (B)

(C) (D)

**11.** The ratio of the corresponding sides of two similar triangles is 1 : 3. The ratio of their corresponding heights is :

(A) 1 : 3 (B) 3 : 1 (C) 1 : 9 (D) 9 : 1

**12.** The areas of two similar triangles are 49 cm^{2} and 64 cm^{2} respectively. The ratio of their corresponding sides is:

(A) 49: 64 (B) 7: 8 (C) 64: 49 (D) None of these

**13.** The areas of two similar triangles are 12 cm^{2} and 48 cm^{2}. If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is:

(A) 4.41 cm (B) 8.4 cm (C) 4.2 cm (D) 0.525 cm

**14.** In the adjoining figure, ABC and DBC are two triangles on the same base BC, AL ^ BC and DM.^ BC. Then, is equal to :

(A) (B)

(C) (D) .

**15.** In the adjoining figure, AD : DC = 2 : 3, then ÐABC is equal to :

(A) 30° (B) 40° (C) 45° (D) 110°

**16.** In DABC, D and E are points on AB and AC respectively such that DE ||BC. If AE = 2 cm, EC = 3 cm and BC = 10 cm, then DE is equal to :

(A) 5 cm (B) 4 cm (C) 15 cm (D) cm

**17.** In the given figure, ÐABC = 90° and BM is a median, AB = 8 cm and BC = 6 cm. Then, length BM is equal to:

(A) 3 cm (B) 4 cm (C) 5 cm (D) 7cm

**18.** If D, E, F are respectively the mid points of the sides BC, CA and AB of DABC and the area of DABC is 24sq. cm, then the area of DDEF is :-

(A) 24 cm^{2} (B) 12 cm^{2} (C) 8 cm^{2} (D) 6 cm^{2}

**19.** In a right angled triangle, if the square of the hypotenuse is twice the product of the other two sides, then one of the angles of the triangle is :-

(A) 15° (B) 30° (C) 45° (D) 60°

**20.** Consider the following statements :

1. If three sides of a triangle are equal to three sides of another triangle, then the triangles are congruent.

2. If three angles of a triangle are respectively equal to three angles of another triangle, then the two triangles are congruent.

Of these statements,

(A) 1 is correct and 2 is false (B) both 1 and 2 are false

(C) both 1 and 2 are correct (D) 1 is false and 2 is correct

**(objective) Exercise**

**ANSWER KEY**

**Exercise-2 (for school/board exams)**

**Subjective type Questions **

**Very Short Answer Type Questions **

**1.** In the given figure, XYBC.

Given that AX = 3 cm, XB = 1.5 cm and BC = 6 cm.

Calculate :

(i) (ii) XY.

**2.** D and E are points on the sides AB and AC respectively of DABC. For each of the following cases, state whether DEBC:

(i) AD = 5.7 cm, BD = 9.5 cm, AE = 3.6 cm and EC = 6 cm

(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 9.6 cm and EC = 2.4 cm.

(iii) AB = 11.7 cm, BD = 5.2 cm, AE = 4.4 cm and AC = 9.9 cm.

(iv) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.

**3.** In DABC, AD is the bisector of ÐA. If BC = 10 cm, BD = 6 cm and AC = 6 cm, find AB.

**4.** AB and CD are two vertical poles of height 6 m and 11 m respectively. If the distance between their feet is 12 m, find the distance between their tops.

**5.** DABC and DPQR are similar triangles such that area (DABC) = 49 cm^{2} and area (DPQR) = 25 cm^{2}. If AB = 5.6 cm, find the length of PQ .

**6.** DABC and DPQR are similar triangles such that area (DABC) = 28 cm^{2} and area (DPQR) = 63 cm^{2}. If PR = 8.4 cm, find the length of AC.

**7.** DABC ~ DDEF. If BC = 4 cm, EF = 5 cm and area (DABC) = 32 cm^{2}, determine the area of DDEF.

**8.** The areas of two similar triangles are 48 cm^{2} and 75 cm^{2} respectively. If the altitude of the first triangle be 3.6 cm, find the corresponding altitude of the other.

**9.** A rectangular field is 40 m long and 30 m broad. Find the length of its diagonal.

**10.** A man goes 15 m due west and then 8 m due north. How far is he from the starting point?

**11.** A ladder 17 m long reaches the window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

**Short Answer Type Questions **

**1.** In the given fig, DEBC.

(i) If AD = 3.6 cm, AB = 9 cm and AE = 2.4 cm, find EC.

(ii) If = and AC = 5.6 cm, find AE.

(iii) If AD = x cm, DB = (x_2) cm, AE = (x+2) cm and EC = (x_1) cm, find the value of x.

**2.** In the given figure, BADC. Show that DOAB ~ DODC. If AB = 4 cm, CD = 3 cm, OC = 5·7 cm and OD = 3·6 cm, find OA and OB.

**3.** In the given figure, ÐABC = 90° and BD ^ AC. If AB = 5·7 cm, BD = 3·8 cm and CD = 5·4 cm, find BC.**4.** In the given figure, DABC ~ DPQR and AM, PN are altitudes, whereas AX and PY are medians. Prove that

**5.** In the given figure, BCDE, area (DABC) = 25 cm^{2}, area (trap. BCED) = 24 cm^{2} and DE = 14 cm. Calculate the length of BC.

**6.** In DABC, ÐC = 90°. If BC = a, AC = b and AB = c, find :

(i) c when a = 8 cm and b = 6 cm.

(ii) a when c = 25 cm and b = 7 cm

(iii) b when c = 13 cm and a = 5 cm

**7.** The sides of a right triangle containing the right angle are (5x) cm and (3x _ 1) cm. If the area of triangle be 60 cm^{2}, calculate the length of the sides of the triangle.

**8.** Find the altitude of an equilateral triangle of side cm.

**9.** In the adjoining figure (not drawn to scale), PS = 4 cm, SR = 2 cm, PT = 3 cm and QT = 5cm.

(i) Show that DPQR ~ DPST. (ii) Calculate ST, if QR = 5·8 cm.

**10.** In the given figure, ABPQ and ACPR. Prove that BCQR.

**11.** In the given figure, AB and DE are perpendicular to BC. If AB = 9 cm, DE = 3 cm and AC = 24 cm, calculate AD.

**12.** In the given figure, DEBC. If DE = 4 cm, BC = 6 cm and area (DADE) = 20 cm^{2}, find the area of DABC.

**13.** A ladder 15 m long reaches a window which is 9 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.

**14.** In the given figure, ABCD is a quadrilateral in which BC = 3 cm, AD = 13 cm, DC = 12 cm and ÐABD = ÐBCD = 90°. Calculate the length of AB.

**15.** In the given figure, ÐPSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm, calculate the length of PR.

**16.** In a rhombus PQRS, side PQ = 17 cm and diagonal PR = 16 cm. Calculate the area of the rhombus.

**17.** From the given figure, find the area of trapezium ABCD.

**18.** In a rhombus ABCD, prove that AC^{2} + BD^{2} = 4AB^{2}.

**19.** A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

**Long Answer Type Questions **

**1.** In the given figure, it is given that ÐABD = ÐCDB = ÐPQB = 90°. If AB = x units, CD = y units and PQ = z units, prove that

**2.** In the adjoining figure, ABCD is a parallelogram, P is a point on side BC and DP when produced meets AB produced at L. Prove that: (i) DP : PL = DC : BL (ii) DL : DP = AL : DC.

**3.** In the given figure, ABCD is a parallelogram, E is a point on BC and the diagonal BD intersects AE at F. Prove that: DF × FE = FB × FA.

**4.** In the adjoining figure, ABCD is a parallelogram in which AB = 16 cm, BC = 10 cm and L is a point on AC such that CL : LA = 2 : 3. If BL produced meets CD at M and AD produced at N, prove that:

(i) DCLB ~ DALN (ii) DCLM ~ DALB

**5.** In the given figure, medians AD and BE of DABC meet at G and DFBE. Prove that

(i) EF = FC (ii) AG : GD = 2 : 1.

**6.** In the given figure, the medians BE and CF of DABC meet at G. Prove that:

(i) DGEF ~ DGBC and therefore, BG = 2GE. (ii) AB × AF = AE × AC.

**7.** In the given figure, DEBC and BD = DC.

(i) Prove that DE bisects ÐADC.

(ii) If AD = 4·5 cm, AE = 3·9 cm and DC = 7-5 cm, find CE.

(iii) Find the ratio AD : DB.

**8.** O is any point inside a DABC. The bisectors of ÐAOB, ÐBOC and ÐCOA meet the sides AB, BC and CA in points D, E and F respectively. Prove that AD**·**BE**·**CF = DB**·**EC**·**FA.

**9.** In the figure, DEBC.

(i) Prove that DADE and DABC are similar.

(ii) Given that AD = BD, calculate DE, if BC = 4.5 cm.

**10.** In the adjoining figure, ABCD is a trapezium in which ABDC and AB = 2 DC. Determine the ratio of the areas of DAOB and DCOD.

**11.** In the adjoining figure, LM is parallel to BC. AB = 6 cm, AL = 2cm and AC = 9 cm. Calculate :

(i) the length of CM.

(ii) the value of

**12.** In the given figure, DEBC and DE : BC = 3 : 5. Calculate the ratio of the areas of DADE and the trapezium BCED.

**13.** In DABC, D and E are mid-points of AB and AC respectively. Find the ratio of the areas of DADE and DABC.

**14.** In a DPQR, L and M are two points on the base QR, such that ÐLPQ = ÐQRP and ÐRPM = ÐRQP. Prove that (i) DPQL ~ DRPM (ii) QL**·**RM = PL**·**PM (iii) PQ^{2} = QL**·**QR

**15.** In the adjoining figure, the medians BD and CE of a DABC meet at G.

Prove that:

(i) DEGD ~ DCGB

(ii) BG = 2 GD from (i) above.

**16.** In the adjoining figure, PQRS is a parallelogram with PQ = 15 cm and RQ = 10 cm. L is a point on RP such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the lengths of PN and RM.

**17.** In DPQR, LMQR and PM : MR = 3 : 4. Calculate:

(i) and then ;

(ii)

(iii)

**18.** In DABC, ÐB = 90° and D is the mid point of BC.

Prove that :

(i) AC^{2} = AD^{2} + 3CD^{2 }

(ii) BC^{2} = 4(AD^{2} _ AB^{2})

**19.** In DABC, if AB = AC and D is a point on BC. Prove that AB^{2} _ AD^{2} = BD × CD.

**Similar Triangle Exercise (X)-CBSE**

**ANSWER KEY**

**Very Short Answer Type Questions **

**1. **(i) (ii) 4 cm **2.** (i) Yes, (ii) No, (iii) No, (iv) Yes **3.** 9 cm **4.** 13 m** 5.** PQ = 4 cm ** 6.** AC = 5.6 cm** 7.** 50 cm^{2} **8.** 4.5 cm **9.** 50 m **10.** 17 m **11.** 8 m

**Short Answer Type Questions **

**1.** (i) 3.6 cm, (ii) 2.1 cm, (iii) x = 4** 2.** OA = 4.8 cm, OB = 7.6 cm **3.** 8.1 cm **5.** 10 cm **6.** (i) 10 cm, (ii) 24 cm, (iii) 12 cm **7.** 15cm, 8 cm, 17cm **8.** 7.5 cm ** 9.** 2.9 cm** 11.** 16 cm **12.** 45 cm^{2} **13.** 21 m **14.** 4 cm **15.** 17 cm **16.** 240 cm^{2} **17.** 14 cm^{2} **19.** 12 m

**Long Answer Type Questions **

**7.** (ii) 6.5 cm, (ii) 3 : 5 **9.** DE = 1.5 cm **10.** 4 : 1 **11.** (i) 6 cm, (ii) ** 12.** 9 : 16** 13.** 1 : 4

**16.** PN = 15 cm, RM = 10 cm **17.** (i) (ii) 3 : 7 (iii) 10 : 7

**Exercise-3 (for school/board exams)**

**Previous years board questions**

**Very Short Answer Type Questions **

**1. **DABC and DDEF are similar, BC = 3 cm, EF = 4 cm and area of DABC = 54 cm^{2}. Determine the area of DDEF. **Delhi-1996 **

**2.** In DABC, CE ^ AB, BD ^ AC and CE & BD intersect at P, considering triangles BEP and CPD, prove that BP × PD = EP × PC. **Delhi-1996C **

**3.** A right triangle has hypotenuse of length q cm and one side of length p cm. If (q _ p ) = 2, express the length of third side of the right triangle in terms of q. ** AI-1996C **

**4.** In the given figure, ABC is a triangle in which AB = AC, D and E are points on the sides AB and AC respectively, such that AD = AE. Show that the points B, C, E and D are concyclic. ** AI-1996C **

**5.** In a DABC, AB = AC and D is a point on side AC, such that BC^{2} = AC × CD. Prove that BD = BC.Your Target is to ** AI-1997 **

**6.** DABC is right angled at B. On the side AC, a point D is taken such that AD = DC and AB = BD. Find the measure of ÐCAB. ** Delhi -1998 **

**7.** In a DABC, P and Q are points on the sides AB and AC respectively such that PQ is parallel to BC. Prove that median AD, drawn from A to BC, bisects PQ. ** AI-1998 **

**8.** Two poles of height 7 m and 12 m stand on a plane ground. If the distance between their feet is 12 m, find the distance between their tips. ** AI-1998C **

**9.** In a DABC, D and E are points on AB & AC respectively such that DE is parallel to BC and AD : DB = 2 : 3. Determine Area (DADE) : Area (DABC). **Foreign-1999 **

**10.** In the given figure, ÐA = ÐB and D & E are points on AC and BC respectively such that AD = BE, show that DE || AB. ** Delhi-1999 **

**11.** In figure, Ð1 = Ð2 and Ð3 = Ð4. Show that PT . QR = PR . ST.** Foreign-2000**

**12. **In figure, LM || NQ and LN || PQ. If MP = MN, find the ratio of the areas of DLMN and DQNP.

**Foreign-2000 **

**13.** ABC is an isosceles triangle right angled at B. Two equilateral triangles BDC and AEC are constructed with side BC and AC. Prove that area of DBCD = area of DACE. **Delhi-2001 **

**14.** The areas of two similar triangles are 81 cm^{2} and 49 cm^{2} respectively. If the altitude of the first triangle is 6.3 cm, find the corresponding altitude of the other.** AI-2001**

**15.** L and M are the mid-points of AB and BC respectively of DABC, right-angled at B. Prove that

4LC^{2} = AB^{2} + 4BC^{2}. ** AI-2001;Foreign-2001 **

**16.** The areas of two similar triangles are 121 cm^{2} and 64 cm^{2} respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other. ** AI-2001 **

**17.** In an equilateral triangle ABC, AD is the altitude drawn from A on side BC. Prove that 3AB^{2 }= 4AD^{2}.

**18.** (i) Prove that the equilateral triangles described on the two sides of a right angled triangle are together equal to the equilateral triangle on the hypotenuse in terms of their areas. **AI-2002**

(ii) P is a point in the interior of DABC. X, Y and Z are points on lines PA, PB and PC respectively such that XY || AB and XZ || AC. Prove that YZ || BC.** AI-2002 ; Delhi-2003 [NCERT]**

(iii) D and E are points on the sides AB and AC respectively of DABC such that DE is parallel to BC and AD : DB = 4 : 5. CD and BE intersect each other at F. Find the ratio of the areas of DDEF and DBCF. ** AI-2000 ; AI-2003**

(iv) P, Q are respective points on sides AB and AC of triangle ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, prove that BC = 3PQ. ** Foreign-2003 **

**19.** D is a point on the side BC of DABC such that ÐADC = ÐBAC. Prove that .

**20.** ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that

**21.** In a DABC, AD ^ BC and . Prove that ABC is a right triangle, right angled at A. **Foreign -2004 **

**22.** In a right angled triangle ABC, ÐA = 90° and AD ^ BC. Prove that AD^{2} = BD × CD.** Delhi -2004C, 2006 **

**23.** In fig., AB || DE and BD || EF. Prove that DC^{2 }= CF × AC.** AI-2004C; Delhi-2007**

**24.** If one diagonal of a trapezium divides the other diagonal in the ratio of 1 : 2, prove that one of the parallel sides is double the other. ** Foreign-2005 **

**25.** In DABC, AD ^ BC, prove that AB^{2} + CD^{2} = AC^{2} + DB^{2}. **Delhi-2005C, AI-2006 [NCERT] **

**26.** Prove that the sum of the squares of the sides of a rhombus is equal to sum of the squares of its diagonals. ** AI-2005C [NCERT] **

**27.** In figure, S and T trisect the side QR of a right triangle PQR. Prove that 8PT^{2} = 3PR^{2} + 5PS^{2}.

**OR**

If BL and CM are medians of a triangle ABC right-angled at A, then prove that 4(BL^{2} + CM^{2}) = 5 BC^{2}.** AI-2006 C; Foreign 2009**

**28.** In the fig, P and Q are points on the sides AB and AC respectively of DABC such that AP = 3.5 cm, PB = 7 cm, AQ = 3 cm and QC = 6 cm. If PQ = 4.5 cm, find BC.** Delhi-2008 **

**29.** In fig, ÐM = ÐN = 46° Express x in terms of a, b and c where a, b and c are lengths of LM, MN and NK respectively.** Delhi-2009 **

**30.** In figure, DABD is a right triangle, right-angled at A and AC ^ BD. Prove that AB^{2} = BC. BD.**AI-2009 **

**31.** In a DABC, DE||BC. If DE = BC and area of DABC = 81 cm^{2}, find the area of DADE.**Foregin-2009**

**Short Answer Type Questions **

**1. **P and Q are points on the sides CA and CB respectively of a DABC right-angled at C. Prove that

AQ^{2} + BP^{2 }= AB^{2} + PQ^{2}. ** Delhi-1996, 2007 **

**2.** ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5 cm and AD = cm, find the length of CE. ** AI -1997 **

**3.** In DABC, if AD is the median, show that AB^{2} + AC^{2} = 2 [AD^{2} + BD^{2}].** Delhi-1997, 98 **

**4.** In the given figure, M is the mid-point of the side CD of parallelogram ABCD. BM, when joined meets AC in L and AD produced in E. Prove that EL = 2BL.** AI -1998; Delhi-1999, AI-2009 **

**5.** ABC is a right triangle, right-angled at C. If p is the length of the perpendicular from C to AB and a, b, c have the usual meaning, then prove that (i) pc = ab (ii) ** Delhi-1998, 98C **

**6.** In an equilateral triangle PQR, the side QR is trisected at S. Prove that 9PS^{2} = 7PQ^{2}.**AI-1998, 98C [NCERT]**

**7.** If the diagonals of a quadrilateral divide each other proportionally, prove that it is trapezium.**Foreign-1999**

**8.** In an isosceles triangle ABC with AB = AC, BD is a perpendicular from B to the side AC. Prove that BD^{2} _ CD^{2} = 2CD . AD.** Foreign-1999 **

**9.** ABC and DBC are two triangles on the same base BC. If AD intersect BC at O. Prove that

**AI-19999C; Delhi-2005 **

**10.** In DABC, ÐA is acute. BD and CE are perpendiculars on AC and AB respectively. Prove that AB × AE = AC × AD. ** AI-2003 **

**11.** Points P and Q are on sides AB and AC of a triangle ABC in such a way that PQ is parallel to side BC. Prove that the median AD drawn from vertex A to side BC bisects the segment PQ.

**12.** If the diagonals of a quadrilateral divide each other proportionally, prove that it is a trapezium.

**OR**

Two Ds' ABC and DBC are on the same base BC and on the same side of BC in which ÐA = ÐD = 90°. If CA and BD meet each other at E, show that AE.EC = BE.ED. ** Delhi-2008 **

**13.** D and E are points on the sides CA and CB respectively of DABC right-angled at C. Prove that AE^{2} + BD^{2} = AB^{2} + DE^{2}.

**OR**

In fig. DB ^ BC, DE ^ AB and AC ^ BC. Prove that . ** AI-2008 **

**14.** E is a point on the side AD produced of a ||^{gm }ABCD and BE intersects CD at F. Show that DABE ~ DCFB. **Foreign-2008 **

**15.** In fig, DABC is right angled at C and DE ^ AB. Prove that DABC ~ DADE and hence find the lengths of AE and DE.

**OR**

In fig, DEFG is a square and ÐBAC = 90°. Show that DE^{2} = BD × EC **Delhi-2009 **

**16.** In fig, AD ^ BC and BD = CD. Prove that 2CA^{2} = 2AB^{2} + BC^{2}.** **

**AI-2009**

**17.** In fig, two triangles ABC and DBC lie on the same side of base BC. P is a point on BC such that PQ||BA and PR||BD. Prove that QR||AD.** Foreign-2009 **

**Long Answer Type Questions **

**1. **In a right triangle ABC, right-angled at C, P and Q are points on the sides CA and CB respectively which divide these sides in the ratio 1 : 2. Prove that ** AI-1996C**

(i) 9AQ^{2} = 9AC^{2} + 4BC^{2} (ii) 9BP^{2 }= 9BC^{2} + 4AC^{2} (iii) 9(AQ^{2} + BP^{2}) = 13AB^{2}.

**2.** The ratio of the areas of similar triangles is equal to the ratio of the squares on the corresponding sides, prove. Using the above theorem, prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.**Delhi-1997C; 2005C; Foreign-2003 **

**3.** Perpendiculars OD, OE and OF are drawn to sides BC, CA and AB respectively from a point O in the interior of a DABC. Prove that :

(i) AF^{2} + BD^{2} + CE^{2} = OA^{2} + OB^{2} + OC^{2} _ OD^{2} _ OE^{2} _ OF^{2}.

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2} + CD^{2} + BF^{2}.** Delhi-1997C, [NCERT] **

**4.** In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares on the other two sides, prove. Using the above theorem, determine the length of AD in terms of b and C. **AI-1997 C **

**5.** If a line is drawn parallel to one side of a triangle, the other two sides are divided in the same ratio, prove. Use this result to prove the following : In the given figure, if ABCD is a trapezium in which AB || DC || EF, then . ** Foreign-1998 **

**6.** State and prove pythagoras theorem. Use the theorem and calculate area (DPMR) from the given figure.

**Delhi-1998C, 2006 **

**7.** In a right-angled triangle, the square of hypotenuse is equal to the sum of the squares of the two sides. Given that ÐB of DABC is an acute angle and AD ^ BC. Prove that AC^{2} = AB^{2} + BC^{2} _ 2BC **.** BD. ** Delhi-1999 **

**8.** In a right triangle, prove that the square on the hypotenuse is equal to the sum of the squares on the other two sides. Using above, solve the following : In quadrilateral ABCD, find the length of CA, if CD ^ DB, AB ^ DB, CD = 6 m, DB = 12 m and AB = 11 m. ** Delhi-2000 **

**9.** Prove that the ratio of the areas of two similar triangles is equal to the squares of their corresponding sides. Using the above, do the following:

In fig. DABC and DPQR are isosceles triangles in which ÐA = ÐP. If find .

**10.** In a right-angled triangle, prove that the square on the hypotenuse is equal to the sum of the squares on the other two sides. Using the above result, find the length of the second diagonal of a rhombus whose side is 5 cm and one of the diagonals is 6 cm. ** AI-2001 **

**11.** In a triangle, if the square on one side is equal to the sum of the squares on the other two sides prove that the angle opposite the first side is a right angle.

Use the above theorem and prove thet following : In triangle ABC, AD ^ BC and BD = 3CD. Prove that 2AB^{2 }= 2AC^{2 }+ BC^{2}.** AI-2003 **

**13.** In a right triangle, prove that the square on hypotenuse is equal to sum of the squares on the other two sides. Using the above result, prove the following : PQR is a right triangle, right angled at Q. If S bisects QR, show that PR^{2} = 4 PS^{2} _ 3 PQ^{2}. ** Delhi-2004C **

**14.** If a line is drawn parallel to one side of a trial prove that the other two sides are divided in the same ratio. Using the above result, prove from fig. that AD = BE if ÐA = ÐB and DE || AB.** AI-2004C **

**15.** Prove that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides. Apply the above theorem on the following : ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1 cm, PB = 3 cm, AQ = 1.5 cm, QC = 4.5 cm, prove that area of DAPQ is one-sixteenth of the area of DABC.** Delhi-2005 **

**16.** If a line is drawn parallel to one side of a triangle, prove that the other two sides are divided in the same ratio. Use the above to prove the following : In the given figure DE || AC and DC || AP. Prove that . ** AI-2005 **

**17.** In a triangle if the square on one side is equal to the sum of squares on the other two sides, prove that the angle opposite to the first side is a right angle. Use the above theorem to prove the following :

In a quadrilateral ABCD, ÐB = 90°. If AD^{2} = AB^{2} + BC^{2} + CD^{2}, prove that ÐACD = 90°.** AI-2005 **

**18.** If a line is drawn parallel to one side of a triangle, to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio. Using the above, prove the following : In figure, DE || BC and BD = CE. Prove that ABC is an isosceles triangle. ** Delhi-2007, 2009 **

**19.** Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Use the above for the following : If the areas of two similar triangles are equal, prove that they are congruent.** AI-2007 **

**20.** Prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides. Using the above result, prove the following :

In a DABC, XY is parallel to BC and it divides DABC into two parts of equal area. Prove that

**ANSWER KEY**

**Similar Triangle Exercise (X)-CBSE**

**Very Short Answer Type Questions **

**2.** 96 cm^{2} ** 3.****6.** 60° **8.** 13 m **9.** 4.25^{ }**12.** 9 : 4** 14.** 4.9 cm **16.** 8.8 cm **18.** (iii) 16 : 81

**28.** 13.5 cm **29.****31.** 36 cm^{2 }

**Short Answer Type Questions **

**2.****15.** AE = , DE =

**Long Answer Type Questions **

**4.****6.** 24 cm^{2 }**8.** 13 cm **9.** 3 : 4 **10.** 8 cm

**Exercise-4 (for Olympiads)**

**Choose The Correct One **

**1.** In a triangle ABC, if AB, BC and AC are the three sides of the triangle, then which of the statements is necessarily true?

(A) AB + BC < AC (B) AB + BC > AC (C) AB + BC = AC (D) AB^{2} + BC^{2} = AC^{2} .

**2.** The sides of a triangle are 12 cm, 8 cm and 6 cm respectively, the triangle is :

(A) acute (B) obtuse (C) right (D) can't be determined

**3.** In an equilateral triangle, the incentre, circumcentre, orthocentre and centroid are:

(A) concylic (B) coincident (C) collinear (D) none of these

**4.** In the adjoining figure D is the midpoint of BC of a DABC. DM and DN are the perpendiculars on AB and AC respectively and DM = DN, then the DABC is :

(A) right angled

(B) isosceles

(C) equilateral

(D) scalene

**5.** Triangle ABC is such that AB = 9 cm, BC = 6 cm, AC = 7.5 cm. Triangle DEF is similar to DABC, If EF = 12 cm then DE is :

(A) 6 cm (B) 16 cm (C) 18 cm (D) 15 cm

**6.** In DABC, AB = 5 cm, AC= 7 cm. If AD is the angle bisector of ÐA. Then BD : CD is:

(A) 25 : 49 (B) 49 : 25 (C) 6 : 1 (D) 5 : 7

**7.** In a DABC, D is the mid-point of BC and E is mid-point of AD, BF passes through E. What is the ratio of AF : FC?

(A) 1 : 1

(B) 1 : 2

(C) 1 : 3

(D) 2 : 3

**8.** In a DABC, AB = AC and AD ^ BC, then :

(A) AB < AD (B) AB > AD (C) AB = AD (D) AB £ AD

**9.** The difference between altitude and base of a right angled triangle is 17 cm and its hypotenuse is 25 cm. What is the sum of the base and altitude of the triangle is ?

(A) 24 cm (B) 31 cm (C) 34 cm (D) can't be determined

**10.** If AB, BC and AC be the three sides of a triangle ABC, which one of the following is true?

(A) AB _ BC = AC (B) (AB _ BC) > AC (C) (AB _ BA) < AC (D) AB^{2} _ BC^{2} = AC^{2}

**11.** In the adjoining figure D, E and F are the mid-points of the sides BC, AC and AB respectively. DDEF is congruent to triangle :

(A) ABC

(B) AEF

(C) CDE , BFD

(D) AFE, BFD and CDE

**12.** In the adjoining figure ÐBAC = 60° and BC = a, AC = b and AB = c, then :

(A) a^{2} = b^{2} + c^{2}

(B) a^{2} = b^{2} + c^{2} _ bc

(C) a^{2} = b^{2} + c^{2} + bc

(D) a^{2} = b^{2} + 2bc

**13.** If the medians of a triangle are equal, then the triangle is:

(A) right angled (B) isosceles (C) equilateral (D) scalene

**14.** The incentre of a triangle is determined by the:

(A) medians (B) angle bisectors

(C) perpendicular bisectors (D) altitudes

**15.** The point of intersection of the angle bisectors of a triangle is :

(A) orthocentre (B) centroid (C) incentre (D) circumcentre

**16.** A triangle PQR is formed by joining the mid-points of the sides of a triangle ABC. 'O' is the circumcentre of DABC, then for DPQR, the point 'O' is :

(A) incentre (B) circumcentre (C) orthocentre (D) centroid

**17.** If AD, BE, CF are the altitudes of DABC whose orthocentre is H, then C is the orthocentre of :

(A) DABH (B) DBDH (C) DABD (D) DBEA

**18.** In an equilateral DABC, if a, b and c denote the lengths of perpendiculars from A, B and C respectively on the opposite sides, then:

(A) a > b > c (B) a > b < c (C) a = b = c (D) a = c ¹ b

**19.** Any two of the four triangles formed by joining the midpoints of the sides of a given triangle are:

(A) congruent (B) equal in area but not congruent

(C) unequal in area and not congruent (D) none of these

**20.** The internal bisectors of ÐB and ÐC of DABC meet at O. If ÐA = 80° then ÐBOC is :

(A) 50° (B) 160° (C) 100° (D) 130°

**21.** The point in the plane of a triangle which is at equal perpendicular distance from the sides of the triangle is :

(A) centroid (B) incentre (C) circumcentre (D) orthocentre

**22.** Incentre of a triangle lies in the interior of :

(A) an isosceles triangle only (B) a right angled triangle only

(C) any equilateral triangle only (D) any triangle

**23.** In a triangle PQR, PQ = 20 cm and PR = 6 cm, the side QR is :

(A) equal to 14 cm (B) less than 14 cm (C) greater than 14 cm (D) none of these

**24.** If ABC is a right angled triangle at B and M, N are the mid-points of AB and BC, then 4 (AN^{2} + CM^{2}) is equal to_

(A) 4AC^{2} (B) 6AC^{2} (C) 5AC^{2} (D) AC^{2}

**25.** ABC is a right angle triangle at A and AD is perpendicular to the hypotenuse. Then is equal to :

(A) (B) (C) (D)

**26.** Let ABC be an equilateral triangle. Let BE ^ CA meeting CA at E, then (AB^{2} + BC^{2} + CA^{2}) is equal to :

(A) 2BE^{2} (B) 3BE^{2} (C) 4BE^{2} (D) 6BE^{2}

**27.** If D, E and F are respectively the mid-points of sides of BC, CA and AB of a DABC. If EF = 3 cm, FD = 4 cm, and AB = 10 cm, then DE, BC and CA respectively will be equal to :

(A) 6, 8 and 20 cm (B) 4, 6 and 8 cm (C) 5, 6 and 8 cm (D) , 9 and 12 cm

**28.** In the right angle triangle ÐC = 90°. AE and BD are two medians of a triangle ABC meeting at F. The ratio of the area of DABF and the quadrilateral FDCE is :

(A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 2 : 3

**29.** The bisector of the exterior ÐA of DABC intersects the side BC produced to D. Here CF is parallel to AD.

(A) (B)

(C) (D) None of these

**30.** The diagonal BD of a quadrilateral ABCD bisects ÐB and ÐD, then:

(A) (B)

(C) AB = AD × BC (D) None of these

**31.** Two right triangles ABC and DBC are drawn on the same hypotenuse BC on the same side of BC. If AC and DB intersects at P, then

(A)

(B) AP × DP = PC × BP

(C) AP × PC = BP × DP

(D) AP × BP = PC × PD

**32.** In figure, ABC is a right triangle, right angled at B. AD and CE are the two medians drawn from A and C respectively. If AC = 5 cm and AD = cm, find the length of CE:

(A) cm (B) 2.5 cm

(C) 5 cm (D) cm

**33.** In a DABC, AB = 10 cm, BC = 12 cm and AC = 14 cm. Find the length of median AD. If G is the centroid, find length of GA :

(A) (B) (C) (D)

**34.** The three sides of a triangles are given. Which one of the following is not a right triangle?

(A) 20, 21, 29 (B) 16, 63, 65

(C) 56, 90, 106 (D) 36, 35, 74

**35.** In the figure AD is the external bisector of ÐEAC, intersects BC produced to D. If AB = 12 cm, AC = 8 cm and BC = 4 cm, find CD.

(A) 10 cm

(B) 6 cm

(C) 8 cm

(D) 9 cm

**36.** In DABC, AB^{2} + AC^{2} = 2500 cm^{2} and median AD = 25 cm, find BC.

(A) 25 cm (B) 40 cm (C) 50 cm (D) 48 cm

**37.** In the given figure, AB = BC and ÐBAC = 15°, AB = 10 cm. Find the area of DABC.

(A) 50 cm^{2}

(B) 40 cm^{2}

(C) 25 cm^{2}

(D) 32 cm^{2}

**38.** In the given figure, if and if AE = 10 cm. Find AB.

(A) 16 cm

(B) 12 cm

(C) 15 cm

(D) 18 cm

**39.** In the figure AD = 12 cm, AB = 20 cm and AE = 10 cm. Find EC.

(A) 14 cm

(B) 10 cm

(C) 8 cm

(D) 15 cm

**40.** In the given fig, BC = AC = AD, ÐEAD = 81°. Find the value of x.

(A) 45°

(B) 54°

(C) 63°

(D) 36°

**41.** What is the ratio of inradius to the circumradius of a right angled triangle?

(A) 1 : 2 (B) 1 : (C) 2 : 5 (D) Can't be determined

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