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**Q.1. In the figure, PS is the bisector of ∠QPR of Δ PQR. Prove that **

**Sol.**

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given that, PS is the angle bisector of ∠QPR.

∠QPS = ∠SPR … (1)

By construction,

∠SPR = ∠PRT (As PS || TR) … (2)

∠QPS = ∠QTR (As PS || TR) … (3)

Using these equations, we obtain

∠PRT = ∠QTR

∴ PT = PR

By construction,

PS || TR

By using basic proportionality theorem for ΔQTR**Q.2. In the figure, D is a point on hypotenuse AC of ΔABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. **

(i) Let us join DB.

We have, DN || CB, DM || AB, and ∠B = 90°

∴ DMBN is a rectangle.

∴ DN = MB and DM = NB

The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90°

⇒ ∠2 + ∠3 = 90° … (1)

In ΔCDM,

∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° … (2)

In ΔDMB,

∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° … (3)

From equation (1) and (2), we obtain

∠1 = ∠3

From equation (1) and (3), we obtain

∠2 = ∠4

In ΔDCM and ΔBDM,

∠1 = ∠3 (Proved above)

∠2 = ∠4 (Proved above)

∴ ΔDCM ∼ ΔBDM (AA similarity criterion)

⇒ DM^{2} = DN × MC

(ii) In right triangle DBN,

∠5 + ∠7 = 90° … (4)

In right triangle DAN,

∠6 + ∠8 = 90° … (5)

D is the foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90°

⇒ ∠5 + ∠6 = 90° … (6)

From equation (4) and (6), we obtain

∠6 = ∠7

From equation (5) and (6), we obtain

∠8 = ∠5

In ΔDNA and ΔBND,

∠6 = ∠7 (Proved above)

∠8 = ∠5 (Proved above)

∴ ΔDNA ∼ ΔBND (AA similarity criterion)

⇒ DN^{2} = AN × NB

⇒ DN^{2} = AN × DM (As NB = DM)**Q.3. In the figure, ABC is a triangle in which ∠ABC >90° and AD ⊥ CB produced. Prove that AC ^{2} = AB^{2} + BC^{2} + 2 BC.BD.**

AB

Applying Pythagoras theorem in ΔACD, we obtain

AC

AC

AC

AC

Thus, we have:

AC

AD

⇒ AD

Applying Pythagoras theorem in ΔADC, we obtain

AD

AB

AB

AC

= AB

AM

Applying Pythagoras theorem in ΔAMC, we obtain

AM

AM

(AM

AD

Using the result, DC = BC/2 , we obtain

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB

= (AD

= (AD

= AD

= AD

(iii) Applying Pythagoras theorem in ΔABM, we obtain

AM

Applying Pythagoras theorem in ΔAMC, we obtain

AM

Adding equations (1) and (2), we obtain

2AM

2AM

2AM

2AM

Let ABCD be a parallelogram.

Let us draw perpendicular DE on extended side AB, and AF on side DC.

Applying Pythagoras theorem in ΔDEA, we obtain

DE

Applying Pythagoras theorem in ΔDEB, we obtain

DE

DE

(DE

DA

Applying Pythagoras theorem in ΔADF, we obtain

AD

Applying Pythagoras theorem in ΔAFC, we obtain

AC

= AF

= AF

= (AF

AC

Since ABCD is a parallelogram,

AB = CD … (iv)

And, BC = AD … (v)

In ΔDEA and ΔADF,

∠DEA = ∠AFD (Both 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common)

∴ ΔEAD ΔFDA (AAS congruence criterion)

⇒ EA = DF … (vi)

Adding equations (i) and (iii), we obtain

DA

DA

BC

[Using equations (iv) and (vi)]

AB

(i) Δ APC ~ Δ DPB

(i) In ΔAPC and ΔDPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

ΔAPC ∼ ΔDPB (By AA similarity criterion)

(ii) We have already proved that

ΔAPC ∼ ΔDPB

We know that the corresponding sides of similar triangles are proportional.

(i) In ΔPAC and ΔPDB,

∠P = ∠P (Common)

∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)

∴ ΔPAC ∼ ΔPDB

(ii)We know that the corresponding sides of similar triangles are proportional.

∴ PA.PB = PC.PD

It is given that,

By using the converse of basic proportionality theorem, we obtain

AD || PC

⇒ ∠BAD = ∠APC (Corresponding angles) … (1)

And, ∠DAC = ∠ACP (Alternate interior angles) … (2)

By construction, we have

AP = AC

⇒ ∠APC = ∠ACP … (3)

On comparing equations (1), (2), and (3), we obtain

∠BAD = ∠APC

⇒ AD is the bisector of the angle BAC.

Let AB be the height of the tip of the fishing rod from the water surface. Let BC be the horizontal distance of the fly from the tip of the fishing rod.

Then, AC is the length of the string.

AC can be found by applying Pythagoras theorem in ΔABC.

AC

AB

AB

AB

Thus, the length of the string out is 3 m.

She pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m

Let the fly be at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

= 2.4 m

In ΔADB,

AB

(1.8 m)

BD

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m

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