For Example:
(i) Any two line segments are similar since length are proportional
(ii) Any two circles are similar since radii are proportional
(iii) Any two squares are similar since corresponding angles are equal and lengths are proportional.
Note: Similar figures are congruent if there is one to one correspondence between the figures.
Two polygons are said to be similar to each other, if:
For Example:
From above we deduce:
Any two triangles are similar, if their
(i) Corresponding angles are equal
(ii) Corresponding sides are proportional
Theorem 1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In ∆ABC, DE  BC.
To prove: AD/DB = AE/EC
Const.: Draw EM ⊥ AD and DN ⊥ AE. Join B to E and C to D.
Proof: In ∆ADE and ∆BDE,
……..(i) [Area of ∆ = 1/2 x base x corresponding altitude]
In ∆ADE and ∆CDE,
∵ DE  BC …[Given ]
∴ ar(∆BDE) = ar(∆CDE)
…[∵ As on the same base and between the same parallel sides are equal in area]
From (i), (ii) and (iii),
Two triangles are similar if either of the following three criterion’s are satisfied:
Results in Similar Triangles based on Similarity Criterion:
Theorem 2. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given: ∆ABC ~ ∆DEF
To prove:
Const.: Draw AM ⊥ BC and DN ⊥ EF.
Proof: In ∆ABC and ∆DEF
…(i) ……[Area of ∆ = 1/2 x base x corresponding altitude]
∵ ∆ABC ~ ∆DEF
…..(ii) …[Sides are proportional]
∠B = ∠E ……..[∵ ∆ABC ~ ∆DEF ]
∠M = ∠N …..[each 90°
∴ ∆ABM ~ ∆DEN …………[AA similarity]
…..(iii) …[Sides are proportional]
From (ii) and (iii), we have:
From (i) and (iv), we have:
Similarly, we can prove that
Results based on Area Theorem:
Note: If the areas of two similar triangles are equal, the triangles are congruent.
Theorem 3: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Given: ∆ABC is a right triangle rightangled at B.
To prove: AB^{2} + BC^{2} = AC^{2}
Const.: Draw BD ⊥ AC
Proof: In ∆s ABC and ADB,
∠A = ∠A …[common]
∠ABC = ∠ADB …[each 90°]
∴ ∆ABC ~ ∆ADB …[AA Similarity]
………[sides are proportional]
⇒ AB^{2} = AC.AD
Now in ∆ABC and ∆BDC
∠C = ∠C …..[common]
∠ABC = ∠BDC ….[each 90°]
∴ ∆ABC ~ ∆BDC …..[AA similarity]
……..[sides are proportional]
BC² = AC.DC …(ii)
On adding (i) and (ii), we get
AB^{2} + BC^{2 }= ACAD + AC.DC
⇒ AB^{2} + BC^{2} = AC.(AD + DC)
AB^{2} + BC^{2} = AC.AC
∴ AB^{2} + BC^{2} = AC^{2}
Theorem 4: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
Given: In ∆ABC, AB^{2} + BC^{2} = AC^{2}
To prove: ∠ABC = 90°
Const.: Draw a right angled ∆DEF in which DE = AB and EF = BC
Proof: In ∆ABC,
AB^{2} + BC^{2} = AC^{2} …(i) [given]
In rt. ∆DEF
DE^{2} + EF^{2} = DF^{2} …[by pythagoras theorem]
AB^{2} + BC^{2} = DF^{2} …..(ii) …[DE = AB, EF = BC]
From (i) and (ii), we get
AC^{2} = DF^{2}
⇒ AC = DF
Now, DE = AB …[by cont]
EF = BC …[by cont]
DF = AC …….[proved above]
∴ ∆DEF ≅ ∆ABC ……[SSS congruence]
∴ ∠DEF = ∠ABC …..[CPCT]
∠DEF = 90° …[by cont]
∴ ∠ABC = 90°
Results based on Pythagoras’ Theorem:
(i) Result on obtuse Triangles.
If ∆ABC is an obtuse angled triangle, obtuse angled at B,
If AD ⊥ CB, then
AC^{2} = AB^{2} + BC^{2 }+ 2 BC.BD
(ii) Result on Acute Triangles.
If ∆ABC is an acute angled triangle, acute angled at B, and AD ⊥ BC, then
AC^{2} = AB^{2 }+ BC^{2} – 2 BD.BC.
116 videos420 docs77 tests


Explore Courses for Class 10 exam
