## EXERCISE 6.6

#### Page No 152:

#### Question 1:

In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that .

#### Answer:

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,

QSSR=QPPT

⇒QSSR=PQPR ∵PT=PR

#### Question 2:

In the given figure, D is a point on hypotenuse AC of ΔABC, DM ⊥ BC and DN ⊥ AB, Prove that:

(i) DM^{2} = DN.MC

(ii) DN^{2} = DM.AN

#### Answer:

(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)

#### Question 3:

In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC^{2} = AB^{2} + BC^{2} + 2BC.BD.

#### Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AB^{2} = AD^{2} + DB^{2} … (1)

Applying Pythagoras theorem in ΔACD, we obtain

AC^{2} = AD^{2 }+ DC^{2}

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

AC^{2} = AD^{2} + DB^{2} + BC^{2} + 2DB × BC

AC^{2} = AB^{2} + BC^{2} + 2DB × BC [Using equation (1)]

#### Question 4:

In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC^{2} = AB^{2} + BC^{2} − 2BC.BD.

#### Answer:

Applying Pythagoras theorem in ΔADB, we obtain

AD^{2} + DB^{2} = AB^{2}

⇒ AD^{2} = AB^{2} − DB^{2 }… (1)

Applying Pythagoras theorem in ΔADC, we obtain

AD^{2} + DC^{2} = AC^{2}

AB^{2} − BD^{2} + DC^{2} = AC^{2} [Using equation (1)]

AB^{2} − BD^{2} + (BC − BD)^{2} = AC^{2}

AC^{2} = AB^{2} − BD^{2} + BC^{2} + BD^{2} −2BC × BD

= AB^{2} + BC^{2} − 2BC × BD

#### Question 5:

In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:

(i)

(ii)

(iii)

#### Answer:

(i) Applying Pythagoras theorem in ΔAMD, we obtain

AM^{2} + MD^{2} = AD^{2 }… (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM^{2} + MC^{2} = AC^{2}

AM^{2} + (MD + DC)^{2} = AC^{2}

(AM^{2} + MD^{2}) + DC^{2} + 2MD.DC = AC^{2}

AD^{2} + DC^{2} + 2MD.DC = AC^{2} [Using equation (1)]

Using the result, , we obtain

(ii) Applying Pythagoras theorem in ΔABM, we obtain

AB^{2} = AM^{2} + MB^{2}

= (AD^{2} − DM^{2}) + MB^{2}

= (AD^{2} − DM^{2}) + (BD − MD)^{2}

= AD^{2} − DM^{2} + BD^{2} + MD^{2} − 2BD × MD

= AD^{2} + BD^{2} − 2BD × MD

(iii)Applying Pythagoras theorem in ΔABM, we obtain

AM^{2} + MB^{2} = AB^{2} … (1)

Applying Pythagoras theorem in ΔAMC, we obtain

AM^{2} + MC^{2} = AC^{2} … (2)

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

2AM^{2} + MB^{2} + MC^{2} = AB^{2} + AC^{2}

2AM^{2} + (BD − DM)^{2} + (MD + DC)^{2} = AB^{2} + AC^{2}

2AM^{2}+BD^{2} + DM^{2} − 2BD.DM + MD^{2} + DC^{2} + 2MD.DC = AB^{2 }+ AC^{2}

2AM^{2} + 2MD^{2} + BD^{2} + DC^{2} + 2MD (− BD + DC) = AB^{2} + AC^{2}

#### Page No 153:

#### Question 6:

Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

#### Answer:

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^{2} + EA^{2} = DA^{2} … (*i*)

Applying Pythagoras theorem in ΔDEB, we obtain

DE^{2} + EB^{2} = DB^{2}

DE^{2} + (EA + AB)^{2} = DB^{2}

(DE^{2} + EA^{2}) + AB^{2} + 2EA × AB = DB^{2}

DA^{2} + AB^{2} + 2EA × AB = DB^{2} … (*ii*)

Applying Pythagoras theorem in ΔADF, we obtain

AD^{2} = AF^{2} + FD^{2}

Applying Pythagoras theorem in ΔAFC, we obtain

AC^{2} = AF^{2} + FC^{2}

= AF^{2} + (DC − FD)^{2}

= AF^{2} + DC^{2 }+ FD^{2} − 2DC × FD

= (AF^{2} + FD^{2}) + DC^{2} − 2DC × FD

AC^{2} = AD^{2} + DC^{2} − 2DC × FD … (*iii*)

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^{2} + AB^{2} + 2EA × AB + AD^{2} + DC^{2} − 2DC × FD = DB^{2} + AC^{2}

DA^{2} + AB^{2} + AD^{2} + DC^{2} + 2EA × AB − 2DC × FD = DB^{2} + AC^{2}

BC^{2} + AB^{2} + AD^{2} + DC^{2} + 2EA × AB − 2AB × EA = DB^{2} + AC^{2}

[Using equations (*iv*) and (*vi*)]

AB^{2} + BC^{2} + CD^{2} + DA^{2} = AC^{2} + BD^{2}

#### Question 7:

In the given figure, two chords AB and CD intersect each other at the point P. prove that:

(i) ΔAPC ∼ ΔDPB

(ii) AP.BP = CP.DP

#### Answer:

Let us join CB.

(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.

∴ AP. PB = PC. DP

#### Question 8:

In the given figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

(i) ΔPAC ∼ ΔPDB

(ii) PA.PB = PC.PD

#### Answer:

(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

#### Question 9:

In the given figure, D is a point on side BC of ΔABC such that. Prove that AD is the bisector of ∠BAC.

#### Answer:

Let us extend BA to P such that AP = AC. Join PC.

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.

#### Question 10:

Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, ho much string does she have out (see Fig. 6.64)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

#### Answer:

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^{2} = AB^{2} + BC^{2}

AB^{2} = (1.8 m)^{2} + (2.4 m)^{2}

AB^{2} = (3.24 + 5.76) m^{2}

AB^{2} = 9.00 m^{2}

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

(1.8 m)^{2} + BD^{2} = (2.4 m)^{2}

BD^{2} = (5.76 − 3.24) m^{2} = 2.52 m^{2}

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m

= 2.787 m

= 2.79 m