# Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.6

### Question 1. In Fig., PS is the bisector of âˆ QPR of Î”PQR. Prove that

Solution:

Given: PS is angle bisector of âˆ QPR.

To prove:

Construction: Draw RTâˆ¥ST to cut OP provided at T.

Proof: since PSâˆ¥TR and PR cuts them, hence we have,

âˆ 1 = âˆ 3           ——[alternate interior angle]  1

âˆ 1 = âˆ 4           ——[corresponding angle]        2

But âˆ 1 = âˆ 2     [given]

From 1 and 2

âˆ´âˆ 3=âˆ 4

In âˆ†PTR,

PT = PR    [sides opposite to equal angle are equal]  3

Now In âˆ†QRT, we have

———-[by B.P.T]

———[from 3]

### Question 2. In Fig., D is a point on hypotenuse AC of Î”ABC, such that BD âŠ¥ AC, DM âŠ¥ BC and

DN âŠ¥ AB. Prove that:

(i) DM2 = DN.MC

(ii) DN2 = DM.AN

Solution:

Given: BD âŠ¥ AC, DM âŠ¥ BC and DN âŠ¥ AB

To prove:

(i) DM2 = DN.MC

(ii) DN2 = DM.AN

Proof: we have ABâŠ¥BC and DMâŠ¥BC

So, AB || DM

Similarly, we have CBâŠ¥AB and DN âŠ¥ AB

Therefore, CB || DN

Hence, BMD DN is a triangle

âˆ´BM=DN or DM=ND

In âˆ†BMD,

âˆ 1+ âˆ 2+ âˆ DMB=180           ———–[angle sum property of triangle]

âˆ 1+ âˆ 2+90=180

âˆ 1+ âˆ 2=180-90

âˆ 1+ âˆ 2=90            ——————-1

Similarly, in âˆ†DMC,

âˆ 3+ âˆ 4=90Â°      _______2

âˆ´ from 1 and 2

âˆ 1+ âˆ 2=âˆ 2+ âˆ 3

âˆ 1=âˆ 3

From 2 and 3

âˆ 3+ âˆ 4=âˆ 2+ âˆ 3

âˆ 4=âˆ 2

Now in âˆ†BMD and âˆ†DMC,

âˆ 1=âˆ 3

âˆ 2=âˆ 4

âˆ´âˆ†BMD~âˆ†DMC           ——-(AA similarity criteria)

DM2=DN*MC

II) proceeding as in (i), we can prove that âˆ†BND~âˆ†DNA

[BN=NA]

DN2 = MN*MC

### Question 3. In Fig., ABC is a triangle in which âˆ ABC > 90Â° and AD âŠ¥ CB produced. Prove that AC2 = AB2 + BC2 + 2 BC.BD.

Given: In âˆ†ABC, âˆ ABC > 90Â° and ADâŠ¥CB provided.

To prove: (AC)2= AB2+BC2+ 2 BC.BD

Proof: In âˆ†ADB by Pythagoras theorem,

### Question 4. In Figure, ABC is a triangle in which âˆ ABC < 90Â° and AD âŠ¥ BC. Prove that AC2 = AB2 + BC2 â€“ 2BC.BD.

Solution:

Given: In âˆ†ABC, âˆ ABC < 90Â° and AD âŠ¥ BC.

To prove: AC2 = AB2 + BC2 â€“ 2*BC.BD

Proof: In triangle ADB, by  Pythagoras theorem

In triangle ADC, by  Pythagoras theorem

AC2=AB2+ BD2+BC2 – 2*BC*BD

AC2 = AB2 + BC2 â€“ 2BC.BD

### Question 5. In Fig., AD is a median of a triangle ABC and AM âŠ¥ BC. Prove that:

(i) AC2 = AD2 + BC.DM + (BC/2)2

(ii) AB2 = AD2 â€“ BC.DM + (BC/2)2

(iii) AC2 + AB2 = 2 AD2 + 1/2 BC2

Given: In âˆ†ABC, AD is a median, AM âŠ¥ BC.

Prove that:

(i) AC2 = AD2 + BC.DM + (BC/2)2

(ii) AB2 = AD2 â€“ BC.DM + (BC/2)2

(iii) AC2 + AB2 = 2 AD2 + 1/2 BC2

Proof: i) In obtuse âˆ†ADC,âˆ D>90Â°

II)In acute triangle ABD,

âˆ D>90Â°

Or

### 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.

Solution:

Given: ||gm ABCD and diagonals BD and AC

Proof: Diagonals of ||gm bisects each other at midpoint

Solution: In âˆ†ABC, BO is median

âˆ´AB2+BC2=2BO2+1/2AC2              ———1

=2.(1/2.BD)2 +1/2AC2+2(1/2BD)2+1/2AC2

=2.BD/42 +1/4AC2+2.BD/42+1/2AC2

=BD2/2+1/2AC2 +BD2/2+1/2AC2

=BD2+AC2

### Question 7. In Fig., two chords AB and CD intersect each other at the point P. Prove that :

(i) Î” APC ~ Î” DPB

(ii) AP.PB = CP.DP

Solution:

Given: chords AB and CD intersect each other at the point P.

To prove: (i) Î” APC ~ Î” DPB           (ii) AP.PB = CP.DP

Proof: In âˆ†ABC and âˆ†DPB,

âˆ 1=âˆ 2            ——-[vertically opposite angle]

âˆ A=âˆ D           ——-[angle in the same segment of circle are equal]

i) âˆ´âˆ†APC~âˆ†DPB    (AA similarity criteria)

ii)

AP*PB=CP*DP

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

(i) Î” PAC ~ Î” PDB

(ii) PA.PB = PC.PD

Solution:

i) In âˆ†PAC and âˆ†PDB,

âˆ P=âˆ P           ———[common]

âˆ 1+âˆ 2=180Â° ——–[linear pair angle]

âˆ 2=180Â°+âˆ 1

âˆ 2+âˆ 3=180Â° ——[sum of opposite angle of a cyclic quadrilateral is 180Â°]

âˆ 3=180Â°-180Â°+âˆ 1

âˆ 3=âˆ 1                   ————2

âˆ 1=âˆ 3 ———–[form 2]

âˆ´âˆ†APC~âˆ†PDB        ———[AA similarity criteria]

ii)

PA*PD = PC*PB

### Question 9. In Fig., D is a point on side BC of Î”ABC such that . Prove that AD is the bisector of âˆ BAC.

Solution:

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

Proof: In âˆ†AEC,

Since AE=AC

âˆ´âˆ 1=âˆ 2             ——[Angle opposite to equal sides of âˆ†are=]

———–(BY CONSTRUCTION)

By converse of BPT

And AC is a transversal then

âˆ 3=âˆ 1       ——-[corresponding angles]

âˆ 4=âˆ 2       ——-[alternate interior angles]

But âˆ 1=âˆ 2      ——–[by construction]

### 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, how much string does she have out (see Fig.)? 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:

In âˆ†ABC,

AC is string

By Pythagoras theorem,

AC2=AB2+BC2

AC2=AB2+BC2

AC2=(1.8)2+(2.4)2

(AC)2=3.24+5.76

AC=âˆš9

AC=3m

âˆ´Length of string she have out=3m

=5cm/sec*12sec

=60cm

=60/100 =0.6m

AP=AC-0.6

=3-06

=2.4 m

Now in âˆ†ABP by Pythagoras theorem,

(AP)2=(AB)2+(PB)2

(2.4)2=(1.8)2+(PB)2

5.76=3.24 + (PB)2

5.76=3.24 + (PB)2

2.52=(PB)2

âˆš2.52= PB

1.59=PB

PD=PB+BD

PD=1.59+1.2

PD=2.79m  ———(approx)

Hence, the horizontal distance of the fly from nazima of the 12 sec in 2.79m.

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