**Maths Class 10
Exercise 6.6**

**1.**** In figure, PS is the bisector
of **∠**QPR of ∆PQR. Prove that QS/SR = PQ/PR. **

**Solution: Given**: PS is the internal bisector of ∠QPR of ∆PQR meeting QR at S.

Therefore, ∠QPS = ∠SPR

**To prove**: QS/SR
= PQ/PR

**Construction**: Draw RT ∥ SP
to cut QP produced at T.

**Proof**: Since PS ∥ TR
and PR is a transversal, then

∠SPR = ∠PRT ………. (i) [Alternate angles]

And ∠QPS = ∠PTR
………. (ii) [Corresponding angles]

But ∠QPS = ∠SPR
[Given]

Therefore, ∠PRT = ∠PTR [From equations (i) and (ii)]

⇒ PT = PR ………. (iii) [Sides opposite to equal angles are
equal]

Now,
in ∆QRT, RT ∥ SP [By construction]

Therefore, QS/SR = PQ/PT [By Basic Proportionality theorem]

⇒ QS/SR = PQ/PR [From equation (iii)] Hence proved.

**2. In figure, D is a point on hypotenuse AC
of ∆ABC, BD **⊥ **AC, DM **⊥** BC
and DN **⊥ **AB. Prove that:**

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

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

**Solution: **Since, AB ⊥ BC
and DM ⊥ BC

⇒ AB ∥ DM

Similarly,
BC ⊥ AB and DN ⊥ AB

⇒ CB ∥ DN

Therefore, quadrilateral
BMDN is a rectangle.

So, BM
= ND

**(i)** In ∆BMD, ∠MBD + ∠BMD
+ ∠BDM = 180°

⇒ ∠MBD + 90° + ∠BDM = 180°

⇒ ∠MBD + ∠BDM
= 90°
……….. (i)

Similarly,
in ∆DMC,
∠MDC + ∠MCD = 90° ………… (ii)

Since
BD ⊥ AC,

Therefore, ∠MDB + ∠MDC = 90° …………. (iii)

From
equations (i) and (iii), we get

⇒ ∠MBD + ∠BDM
= ∠MDB + ∠MDC

⇒ ∠MBD
= ∠MDC

From
equations (ii) and (iii), we get

⇒∠MDC + ∠MCD = ∠MDB
+ ∠MDC

⇒ ∠MCD
= ∠ MDB

Thus,
in ∆BMD and ∆DMC,

∠MBD = ∠MDC and ∠MCD
= ∠ MDB

Therefore, ∆BMD
~ ∆DMC

⇒ BM/DM = MD/MC

⇒ DN/DM = DM/MC [BM = DN]

⇒ DM^{2}
= DN . MC

**(ii)** As proved in (i), we can prove that ∆BND ~
∆DNA.

⇒ BN/DN = DN/AN

⇒ DM/DN = DN/AN [BN = DM]

⇒ DN^{2}
= DM . AN

**3. In figure, ABC is a triangle in which **∠**ABC > 90° and
AD **⊥ **CB produced. Prove that:**

**AC ^{2} = AB^{2} + BC^{2 }+
2BC . BD**

**Solution: Given**: ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB
produced.

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

**Proof**: Since
∆ADB is a right triangle, right angled at D,

Therefore,
AB^{2} = AD^{2 }+ DB^{2 }……… (i) [By Pythagoras theorem]

Again, ∆ADC is a right triangle, right angled at D.

Therefore,
AC^{2} = AD^{2} + DC^{2 }[by
Pythagoras theorem]

⇒ AC^{2} =
AD^{2} + (DB + BC)^{2 }^{ }[Since DC = DB + BC]

⇒ AC^{2} =
AD^{2} + DB^{2} + BC^{2} + 2DB . BC

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

⇒ AC^{2} = AB^{2} + BC^{2}+ 2DB.BC [Using equation (i)] Hence proved.

**4. In figure, ABC is a triangle in which **∠**ABC < ****90°**** and
AD **⊥ **BC. Prove that:**

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

**– 2BC . BD**

**Solution: Given**: ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC.

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

**Proof**: Since ∆ADB is a right triangle, right angled
at D.

Therefore,
AB^{2} = AD^{2} + BD^{2 } ……… (i) [By Pythagoras theorem]

Again,
∆ADC
is a right triangle, right angled at D.

Therefore,
AC^{2} = AD^{2} + DC^{2 } [By Pythagoras theorem]

⇒ AC^{2} = AD^{2} + (BC – BD)^{2 } [Since DC = BC – BD]

⇒ AC^{2} = AD^{2} + BC^{2}
+ BD^{2 }– 2BC . BD

⇒ AC^{2} = (AD^{2} + BD^{2})
+ BC^{2 }– 2 BC . BD

⇒ AC^{2} = AB^{2} + BC^{2 }–
2BC . BD [Using equation (i)] Hence proved.

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

**(i) AC ^{2} = AD^{2} + BC . DM
+ (BC/2)^{2} **

**(ii) AB ^{2} = AD^{2} **

**– BC . DM + (BC/2)**

^{2}**(iii) ****AC ^{2} +
AB^{2} = 2AD^{2} + ½ BC^{2} **

**Solution: **Since ∠AMD = 90°, therefore, ∠ADM < 90° and ∠ADC > 90°

Thus, ∠ADM is an acute angle and ∠ADC is an obtuse angle.

**(i)** In ∆ADC, ∠ADC is an obtuse angle.

Therefore, AC^{2} = AD^{2}
+ DC^{2} + 2DC . DM

⇒ AC^{2}
= AD^{2} + (BC/2)^{2} + 2BC/2 . DM [Since DC = BC/2]

⇒ AC^{2}
= AD^{2} + (BC/2)^{2} + BC . DM

⇒ AC^{2}
= AD^{2} + BC . DM + (BC/2)^{2
} ………….. (i)

**(ii)** In ∆ABD, ∠ADM
is an acute angle.

Therefore,
AB^{2} = AD^{2} + BD^{2} – 2BD . DM

⇒ AB^{2} = AD^{2} + (BC/2)^{2}
– 2BC/2 . DM [Since BD = BC/2]

⇒ AB^{2} = AD^{2} + (BC/2)^{2}
– BC . DM

⇒ AB^{2} = AD^{2} – BC . DM
+ (BC/2)^{2}^{
} ………….. (ii)

**(iii)** Adding equations (i) and (ii),
we get

AC^{2}
+ AB^{2 }= 2AD^{2} + ½ BC^{2}

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

**Solution: Given: **A
parallelogram ABCD whose diagonals AC and BD intersect at point O.

**To prove: **AB^{2} + BC^{2} + AD^{2} + CD^{2} = AC^{2} + BD^{2}

If AD is a median of ∆ABC, then

AC^{2}
+ AB^{2 }= 2AD^{2} + ½ BC^{2} [See Q.5 (iii)]

Since the
diagonals of a parallelogram bisect each other, therefore, BO and DO are
medians of triangles ABC and ADC respectively.

Therefore,
AB^{2} + BC^{2} = 2BO^{2} + ½ AC^{2} ………. (i)

And AD^{2} + CD^{2} = 2DO^{2} + ½ AC^{2} ………. (ii)

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

AB^{2} + BC^{2} + AD^{2} + CD^{2} = 2(BO^{2} + DO^{2}) + AC^{2}

⇒ AB^{2} + BC^{2} + AD^{2} + CD^{2} = 2(¼ BD^{2} + ¼ BD^{2}) +
AC^{2 }[DO = ½ BD]

⇒ AB^{2} + BC^{2} + AD^{2} + CD^{2} = AC^{2} + BD^{2} Hence proved.

**7. In figure, two chords AB and CD intersect
each other at the point P. Prove that:**

**(i) ****∆****APC ****~ ****∆DPB**

**(ii) AP . PB = CP . DP**

**Solution:
(i)** In the
triangles APC and DPB,

∠APC = ∠DPB [Vertically opposite angles]

∠CAP = ∠BDP [Angles in the same segment of a
circle are equal]

Therefore, ∆APC ~ ∆DPB [By AAA criterion of
similarity]

**(ii)** Since ∆APC ~ ∆DPB

Therefore, AP/DP = CP/PB

⇒ AP . PB = CP . DP

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

**(i) ****∆****PAC ****~ ****∆PDB**

**(ii) PA . PB = PC . PD**

**Solution: (i)** In the ∆PAC
and ∆PDB, we have

∠APC = ∠DPB [Common]

∠BDP = 180°– ∠BAC

= 180°– (180° –∠PAC)
= ∠PAC [Since ∠BAC = 180°– ∠PAC]

⇒ ∠PAC
= ∠BDP

Therefore, ∆APC ~ ∆PDB [By AAA criterion of similarity]

** **

**(ii)** Since ∆APC ~ ∆PDB

Therefore, PA/PD = PC/PB

⇒ PA . PB = PC . PD

**9. In figure, D is a point on side BC of ****∆****ABC
such that BD/CD = AB/AC. Prove that AD is the bisector of ****∠****BAC.**

**Solution: Given**: ABC is a triangle and D is a point on BC such that BD/CD = AB/AC.

**To prove**: AD is the internal bisector of ∠BAC.

**Construction**: Produce BA to E such that AE = AC. Join CE.

**Proof**: In ∆AEC, we have

AE = AC

Therefore, ∠AEC
= ∠ACE ………. (i) [Angles opposite to equal side of a
triangle are equal]

Now, BD/CD = AB/AC [Given]

⇒ BD/CD = AB/AE [Since, AC = AE, by construction]

Therefopre, DA ∥ CE [By converse of Basic Proportionality theorem]

Now,
since CA is a transversal.

Therefore, ∠BAD = ∠AEC
………. (ii) [Corresponding angles]

And ∠DAC = ∠ACE
………. (iii) [Alternate angles]

Also ∠AEC = ∠ACE
[From equation (i)]

Hence, ∠BAD = ∠DAC
[From equations (ii) and (iii)]

Thus,
AD bisects ∠BAC internally.

**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, how much string does she have out (see figure)? 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?**

**Solution: (i)**

To
find the length of AC, we have

AC^{2 }= AB^{2} + BC^{2} ^{ }[By
Pythagoras theorem]

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

⇒ AC^{2} = 5.76 +
3.24 = 9.00

⇒ AC = 3 m

Therefore, the length of the string she has out = 3 m

**(ii)** The length of the string pulled at the rate
of 5 cm/sec in 12 seconds

=
(5 × 12) cm = 60 cm = 0.60 m

Therefore, the remaining string left out = 3 – 0.6 = 2.4 m

To find the length of PB, we have

PB^{2} = PC^{2} – BC^{2 }[By Pythagoras
theorem]

=
(2.4)^{2} – (1.8)^{2}

= 5.76 – 3.24

= 2.52

⇒ PB = √2.52 = 1.59 (approx.)

Hence,
the horizontal distance of the fly from Nazima after 12 seconds

=
1.59 + 1.2 = 2.79 m (approx.)