# ICSE Solutions for Selina Concise Chapter 13 Pythagoras Theorem Class 9 Maths

**Exercise 13(A)**

**1. A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.**

**Answer**

The pictorial representation of the given problem is given below,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Here, AB is the hypotenuse. Therefore applying the Pythagoras theorem we get,

13^{2} = 5^{2} + AC^{2}

⇒ 169 = 25 + AC^{2}

⇒ AC^{2} = 169 – 25

⇒ AC^{2 }= 144

⇒ AC = 12

Therefore, the distance of the other end of the ladder from the ground is 12m.**2. A man goes 40 m due north and then 50 m due west. Find his distance from the starting point.**

**Answer**

Here, we need to measure the distance AB as shown in the figure below,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.Therefore, in this case

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

⇒ AB^{2} = 50^{2} + 40^{2}

⇒ AB^{2} = 2500 + 1600

⇒ AB^{2 }= 4100

⇒ AB = 64.03

Therefore, the required distance is 64.03 m.

**3. In the figure: PSQ = 90**

^{o}, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.**Answer**

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

PQ^{2} = PS^{2} + QS^{2}

⇒ 10^{2} = PS^{2} + 6^{2}

⇒ PS^{2} = 100 – 36

⇒ PS = 8

Now, we consider the ∆PRS and applying Pythagoras theorem we get,

PR^{2} = RS^{2} + PS^{2}

⇒ PR^{2} = 15^{2} + 8^{2}

⇒ PR = 17

The length of PR 17Cm

First, we consider the Î”PQS and applying Pythagoras theorem we get,

**4. The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ABD = BCD = 90**

^{o}. Calculate the length of AB.**Answer**

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the Î”BCD and applying Pythagoras theorem we get,

DB^{2} = DC^{2} + BC^{2}

⇒ DB^{ 2} = 12^{2} + 3^{2}

⇒ DB^{ 2} = 144 + 9

⇒ DB^{2} = 153

Now, we consider the ∆ABD and applying Pythagoras theorem we get,

DA^{ 2} = DB^{ 2} + BA^{2}

⇒ 13^{2} = 153 + BA^{2}

⇒ BA = 4

The length of AB is 4 cm.

**5. AD is drawn perpendicular to base BC of an equilateral triangle ABC. Given BC = 10 cm, find the length of AD, correct to 1 place of decimal.**

**Answer**

Since ABC is an equilateral triangle therefore, all the sides of the triangle are of same measure and the perpendicular AD will divide BC in two equal parts..

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Here, we consider the and applying Pythagoras theorem we get

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

⇒ AD^{2} = 10^{2} - 5^{2} **[Given, BC = 10 CM = AB, BD = (1/2)BC]**

⇒ AD^{2} = 100 – 25

⇒ AD^{2} = 75

⇒ AD = 8.7

Therefore, the length of AD is 8.7 cm

**6. In triangle ABC, given below, AB = 8 cm, BC = 6 cm and AC= 3 cm. Calculate the length of OC.**

**Answer**

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the and applying Pythagoras theorem we get,

AB^{2} = AO^{2} + OB^{2}

⇒ AO^{2} = AB^{2} – OB^{2}

⇒ AO^{2} = AB^{2} – (BC + OC)^{2 } **[Let, OC = x]**

⇒ AO^{2} = AB^{2} – (BC + x)^{2} **…(i)**

First, we consider the ∆ACO, and applying pythagoras theorem we get,

AC^{2} = AO^{2}+ x^{2}

⇒ AO^{2} = AC^{2} – x^{2} **…(ii)**

Now, from (i) and (ii),

AB^{2} – (BC + x)^{2} = AC^{2} – x^{2}

⇒ 8^{2} – (6+x)^{2} = 3^{2} – x^{2} **[Given, AB = 8cm, BC = 8cm and AC = 3cm]**

⇒ x = 1(7/12)cm

Therefore, the length of OC will be 1(7/12)cm

**7. In triangle ABC,**

**AB = AC = x, BC = 10 cm and the area of the triangle is 60 cm ^{2}. Find x.**

**Answer**

Here, the diagram will be,

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since, ABC is an isosceles triangle, therefore perpendicular from vertex will cut the base in two equal

First, we consider the Î”ABD, and applying Pythagoras theorem we get,

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

⇒ AD^{2} = x^{2} - 5^{2}

⇒ AD^{2} = x^{2} – 25

⇒ AD = √(x^{2} – 25) **…(i)**

Now,

Area = 60

(1/2)×10×AD = 60

⇒ (1/2)×10× √(x^{2} – 25) = 60

⇒ x = 13

Therefore, x is 13cm.

**8. If the sides of triangle are in the ratio 1:1, show that is a right-angled triangle.**

**Answer**

Let, the sides of the triangle be, x, √2 x and x

Now, x^{2} + x^{2} = 2x^{2} = (√2x)^{2} **…(i)**

Here, in (i) it is shown that, square of one side of the given triangle is equal to the addition of square of other two sides. This is nothing but Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Therefore, the given triangle is a right angled triangle.

**9. Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m; find the distance between their tips.**

**Answer**

The diagram of the given problem is given below,

We have Pythagoras theorem which states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Here, 11 - 6 = 5 m

**(Since DC is perpendicular to BC)**

Base = 12m

Applying Pythagoras theorem we get,

Hypotenuse

^{2}= 5

^{2}+ 12

^{2}

⇒ h

^{2}= 25 + 144

⇒ h

^{2}= 169

⇒ h = 13

Therefore, the distance between the tips will be 13m.

**10. In the given figure, AB||CD, AB = 7 cm, BD = 25 cm and CD = 17 cm; find the length of side BC.**

**Answer**

Take M be the point on CD such that AB = DM.

So DM = 7cm and MC = 10 cm

Join points B and M to form the line segment BM.

So BM || AD also BM = AD.

In right - angled Î”BAD

BD

^{2}= AD

^{2}+BA

^{2}

⇒ (25)

^{2}= AD

^{2}+ (7)

^{2}

⇒ AD

^{2}= (25)

^{2}- (7)

^{2}

⇒ AD

^{2}= 576

⇒ AD = 24

In right - angled △CMB

CB

^{2}= CM

^{2}+ MB

^{2}

⇒ CB

^{2}= (10)

^{2}+ (24)

^{2}

**[MB = AD]**

⇒ CB

^{2}= 100 + 576

⇒ CB

^{2}= 676

⇒ CB = 26CM

**11. In the given figure, ∠B = 90**

^{°}, XY || BC, AB = 12cm, AY = 8cm and AX: XB = 1: 2 = AY: YC. Find the lengths of AC and BC.**Answer**

Given that AX:XB = 1:2 = AY:YC.

Let x be the common multiple for which this proportion gets satisfied.

So, AX = 1x and XB = 2x

AX + XB = 1x + 2x = 3x

⇒ AB = 3x

⇒ 12 = 3x

⇒ x = 4

AX = 1x = 4 and XB = 2x = 2×4 = 8

Similarly,

AY = 1y and YC = 2y

⇒ AY = 8 **(given)**

⇒ 8 = y

∴ YC = 2y = 2×8 = 16

∴ AC = AY + YC = 8 + 16 = 24 cm

∆ABC is a right angled triangle. **(Given)**

∴ By Pythagoras Theorem, we get

⇒ AB^{2} + BC^{2} = AC^{2}

⇒ BC^{2 }= AC^{2} – AB^{2}

⇒ BC^{2 }= (24)^{2} – (12)^{2}

⇒ BC^{2 }= 576 – 144

⇒ BC^{2 }= 432

⇒ BC = 12√3 cm

∴AC = 24cm and BC = 12√3 cm

**12. In Î”ABC, ∠B = 90°. Find the sides of the triangle, if:**

**(i) AB = (x – 3) cm, BC = (x + 4) cm and AC = (x + 6) cm**

**(ii) AB = x cm, BC = (4x + 4) cm and AC = (4x + 5) cm**

**Answer**

**(i)**In right - angled Î”ABC

AC

^{2}= AB

^{2}+ BC

^{2}

⇒ (x+6)

^{2}= (x-3)

^{2}+(x+4)

^{2}

⇒ (x

^{2}+ 12x +36) = (x

^{2}- 6x +9) + (x

^{2}+ 8x + 16)

⇒ x

^{2}- 10x - 11 = 0

⇒ (x - 11)(x+1) = 0

⇒ x = 11 or x = -1

But length of the side of a triangle can not be negative.

⇒ x = 11cm

∴ AB = (x -3) = (11-3) = 8cm

BC = (x+4) = (11+4) = 15cm

AC = (x+6) = (11+6) = 17cm

**(ii)**In right - angled Î”ABC

AC

^{2}= AB

^{2}+ BC

^{2}

⇒ (4x +5)

^{2}= (x)

^{2}+ (4x+4)

^{2}

⇒ (16x

^{2}+ 40x + 25) = (x

^{2}) + (16x

^{2}+ 32x + 16)

⇒ x

^{2}-8x-9 = 0

⇒ (x - 9)(x+1) = 0

⇒ x = 9 or x = -1

But length of the side of a triangle can not be negative.

x = 9 cm

∴ AB = x = 9 cm

BC = (4x +4) = (36+4) = 40cm

AC = (4x + 5) = (36+5) = 41cm

**Exercise 13(B)**

**1.**

**In the figure, given below, AD BC. Prove that: c**^{2}= a^{2}+ b^{2}– 2ax.**Answer**

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

First, we consider the Î”ACD and applying Pythagoras theorem we get,

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

⇒ c^{2} = h^{2} +(a-x)^{2}

⇒ h^{2} = c^{2} - (a-x)^{2} **...(i) **

First, we consider the Î”ACD and applying Pythagoras theorem we get,

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

⇒ b^{2} = h^{2} + x^{2}

⇒ h^{2} = b^{2} -x^{2} **...(ii)**

From (i) and (ii) we get,

c^{2} - (a-x)^{2} = b^{2} - x^{2}

⇒ c^{2} -a^{2} -x^{2} +2ax = b^{2} - x^{2}

⇒ c^{2} = a^{2} + b^{2} -2ax

**2. In equilateral Î” ABC, AD…. BC and BC = x cm. Find, in terms of x, the length of AD.**

**Answer**

AC

^{2}= AD

^{2}+ DC

^{2}

⇒ (x)

^{2}= AD

^{2}+ (x/2)

^{2}

⇒ AD

^{2}= (x)

^{2}- (x/2)

^{2}

⇒ AD

^{2}= (x/2)

^{2}

⇒ AD = (x/2)cm

**3. ABC is a triangle, right-angled at B. M is a point on BC. Prove that:**

**AM ^{2} + BC^{2} = AC^{2} + BM^{2}.**

**Answer**

The pictorial form of the given problem is as follows,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.First, we consider the Î”ABM and applying Pythagoras theorem we get,

AM

^{2}= AB

^{2}+ BM

^{2}

⇒ AB

^{2}= AM

^{2}- BM

^{2}

**...(i)**

Now, we consider the Î”ABC and applying Pythagoras theorem we get,

AC

^{2}= AB

^{2}+ BC

^{2}

⇒ AB

^{2}= AC

^{2}- BC

^{2}

**...(ii)**

From (i) and (ii) we get,

AM

^{2}- BM

^{2}= AC

^{2}- BC

^{2}

⇒ AM

^{2}+ BC

^{2}= AC

^{2}+ BM

^{2}

Hence, Proved

**4. M and N are the mid-points of the sides QR and PQ respectively of a PQR, right-angled at . Prove that:**

**(i) PM ^{2} + RN^{2} = 5 MN^{2}**

**(ii) 4 PM ^{2} = 4 PQ^{2} + QR^{2}**

**(iii) 4 RN ^{2} = PQ^{2} + 4 QR^{2}**

**(iv) 4 (PM ^{2} + RN^{2}) = 5 PR^{2}**

**Answer**

We draw, PM,MN,NR

Since, M and N are the mid-points of the sides QR and PQ respectively, therefore, PN=NQ, QM=RM

(i) First, we consider the Î”PQM, and applying Pythagoras theorem we get,

PM^{2} = PQ^{2} + MQ^{2}

= (PN + NQ)^{2} + MQ^{2}

= PN^{2} + NQ^{2} + 2PN.NQ + MQ^{2}

= MN^{2} +PN^{2} + 2PN.NQ **[From, Î”MNQ, MN ^{2} = NQ^{2} + MQ^{2} ] ...(i)**

Now, we consider the Î”RNQ, and applying Pythagoras theorem we get,

RN

^{2}= NQ

^{2}+ RQ

^{2}

= NQ

^{2}+ (QM+RM)

^{2}

= NQ

^{2}+ QM

^{2}+ RM

^{2}+ 2QM.RM

= MN

^{2}+ RM

^{2}+ 2QM.RM

**...(ii)**

Adding (i) and (ii) we get,

PM

^{2}+ RN

^{2}= MN

^{2}+ PN

^{2}+2PN.NQ +MN

^{2}+RM

^{2}+ 2QM.RM

⇒ PM

^{2}+RM

^{2}= 2MN

^{2}+PN

^{2}+ RM

^{2}+ 2PN.NQ + 2QM.RM

⇒ PM

^{2}+RN

^{2}= 2MN

^{2}+ NQ

^{2}+ QM

^{2}+ 2(QN

^{2}) +2(QM

^{2})

⇒ PM

^{2}+RN

^{2}= 2MN

^{2}+ MN

^{2}+2MN

^{2}

⇒ PM

^{2}+RN

^{2}= 5MN

^{2}

Hence, Proved

**(ii)** We consider the Î”PQM, and applying Pythagoras theorem we get,

PM^{2} = PQ^{2} +MQ^{2}

⇒ 4PM^{2} = 4PQ^{2} + 4MQ^{2} **[Multiplying both sides by 4]**

⇒ 4PM^{2} = 4PQ^{2} + 4[(1/2)QR]^{2} **[MQ = (1/2)QR]**

⇒ 4PM^{2} = 4PQ^{2} + 4.(1/4)QR^{2}

⇒ 4PM^{2} = 4PM^{2} + QR^{2}

Hence, Proved.

(iii) We consider the Î”RQN, and applying Pythagoras theorem we get,

RN^{2} = NQ^{2} + RQ^{2}

⇒ 4RN^{2} = 4NQ^{2} + 4QR^{2} **[Multiplying both sides by 4]**

⇒ 4RN^{2} = 4QR^{2} + 4[(1/2)PQ]^{2} **[NQ = (1/2)PQ]**

⇒ 4RN^{2} = 4QR^{2} + 4. (1/4)PQ^{2}

⇒ 4RN^{2} = PQ^{2} + 4QR^{2}

Hence, Proved

(iv) First, we consider the Î”PQM, and applying Pythagoras theorem we get

PM^{2} = PQ^{2} + MQ^{2}

= (PN+ NQ)^{2} + MQ^{2}

= PN^{2} + NQ^{2} + 2PN. NQ + MQ^{2}

= MN^{2} + PN^{2} + 2PN.NQ **[From, Î”MNQ, MN ^{2} = NQ^{2} + MQ^{2} ] ...(i)**

Now, we consider the Î”RNQ, and applying Pythagoras theorem we get,

RN

^{2}= NQ

^{2}+ RQ

^{2}

= NQ

^{2}+(QM+RM)

^{2}

= NQ

^{2}+ QM

^{2}+ RM

^{2}+ 2QM. RM

= MN

^{2 }+ RM

^{2}+ 2QM.RM

**...(ii)**

Adding (i) and (ii) we get,

PM

^{2}+ RN

^{2}= MN

^{2}+ PN

^{2}+ 2PN.NQ+MN

^{2}+ RM

^{2}+2QM.RM

⇒ PM

^{2}+ RN

^{2}= 2MN

^{2}+PN

^{2}+RM

^{2}+ 2PN.NQ+2QM.RM

⇒ PM

^{2}+ RN

^{2}= 2MN

^{2}NQ

^{2}+ QM

^{2}+ 2(QN

^{2})+2(QM

^{2})

⇒ PM

^{2}+ RN

^{2}= 2MN

^{2}+ MN

^{2}+ 2MN

^{2}

⇒ PM

^{2}+ RN

^{2}= 5MN

^{2}

⇒ 4(PM

^{2}+ RN

^{2})= 4.5 (NQ

^{2}+ MQ

^{2})

**5. In triangle ABC, ∠B = 90**

^{o}and D is the mid-point of BC. Prove that: AC^{2}= AD^{2}+ 3CD^{2}.**Answer**

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

In triangle ABC, ∠B = 90° and D is the mid-point of BC. Join AD . Therefore, BD = DC

First, we consider the Î”ADB, and applying Pythagoras theorem we get,

AD

^{2}= AB

^{2}+ BD

^{2}

⇒ AB

^{2}= AD

^{2}- BC

^{2}

**...(i)**

Similarly, we get rom rt. angle triangles ABC we get,

AC

^{2}= AB

^{2}+ BC

^{2}

⇒ AB

^{2}= AC

^{2}- BC

^{2}

**...(ii)**

From (i) and (ii),

AC

^{2}- BC

^{2}= AD

^{2}- BD

^{2}

⇒ AC

^{2}=AD

^{2}- BD

^{2}+ BC

^{2}

⇒ AC

^{2}= AD

^{2}- CD

^{2}+ 4CD

^{2}

**[BD = CD = (1/2)BC]**

⇒ AC

^{2}= AD

^{2}+ 3CD

^{2}

Hence, proved

**6. In a rectangle ABCD, prove that:**

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

**Answer**

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Since, ABCD is a rectangle angles A,B,C and D are rt. angles.

First, we consider the Î”ACD, and applying Pythagoras theorem we get,

AC

^{2}= DA

^{2}+ CD

^{2}

**...(i)**

Similarly, we get from rt. angle triangle BDC we get,

BD

^{2}= BC

^{2}+ CD

^{2}

= BC

^{2}+ AB

^{2}

**[In a rectangle, opposite sides are equal, CD = AB] ...(ii)**

Adding (i) and (ii),

AC

^{2}+ BD

^{2}= AB

^{2}+ BC

^{2}+ CD

^{2}+ DA

^{2}

Hence, proved.

**7. In a quadrilateral ABCD, ∠B = 90**

^{0}and D = 90^{0}. Prove that: 2AC^{2}– AB^{2}= BC^{2}+ CD^{2}+ DA^{2}**Answer**

In Î”ABC using Pythagoras theorem,

AC

^{2}= AB

^{2}+ BC

^{2}

⇒ AB

^{2}= AC

^{2}- BC

^{2}

**...(i)**

In Î”ADC, using Pythagoras theorem,

AC

^{2}= AD

^{2}+ DC

^{2}

LHS = 2AC

^{2}- AB

^{2}

= 2AC

^{2}- (AC

^{2}- BC

^{2})

**[from(i)]**

= 2AC

^{2}- AC

^{2}+ BC

^{2}

= AC

^{2}+ BC

^{2}

= AD

^{2}+ DC

^{2}+ BC

^{2}

**[from (ii)]**

= RHS

**8. O is any point inside a rectangle ABCD. Prove that: OB**

^{2}+ OD^{2}= OC^{2}+ OA^{2}.**Answer**

Draw rectangle ABCD with arbitrary point O within it, and then draw lines OA, OB, OC, OD. Then draw lines from point O perpendicular to the sides: OE, OF, OG, OH.

Using Pythagorean theorem we have from the above diagram:

OA^{2 }= AH^{2 }+ OH^{2 }= AH^{2 }+ AE^{2}

⇒ OC^{2 }= CG^{2 }+ OG^{2 }= EB^{2 }+ HD^{2}

⇒ OB^{2 }= EO^{2 }+ BE^{2 }= AH^{2 }+ BE^{2}

⇒ OD^{2 }= HD^{2 }+ OH^{2 }= HD^{2 }+ AE^{2}

Adding these equalities we get:

OA^{2 }+ OC^{2 }= AH^{2 }+ HD^{2 }+ AE^{2 }+ EB^{2}

⇒ OB^{2 }+ OD^{2 }= AH^{2 }+ HD^{2 }+ AE^{2 }+ EB^{2}

From which we prove that for any point within the rectangle there is the relation

OA^{2 }+ OC^{2 }= OB^{2 }+ OD^{2}

Hence, Proved.

**9. In the following figure, OP, OQ and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC. Prove that:**

**AR ^{2} + BP^{2} + CQ^{2} = AQ^{2} + CP^{2} + BR^{2}**

**Answer**

Here, we first need to join OA, OB, and OC after which the figure becomes as follows,

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides. First, we consider the Î”ARO and applying Pythagoras theorem we get,AO

^{2}= AR

^{2}+ OR

^{2}

⇒ AR

^{2}= AO

^{2}- OR

^{2}

**...(i)**

similarly, from triangles, BPO, COQ, AOQ, CPO and BRO we get the following results,

BP

^{2}= BO

^{2}- OP

^{2}

**...(ii)**

CQ

^{2}= OC

^{2}- OQ

^{2}

**...(iii)**

AQ

^{2}= AO

^{2}- OQ

^{2}

**...(iv)**

CP

^{2}= OC

^{2}- OP

^{2}

**...(v)**

BR

^{2}= OB

^{2}- OR

^{2}

**...(vi)**

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

AR

^{2}+ BP

^{2}+ CQ

^{2}= AO

^{2}- OR

^{2}+ BO

^{2}- OP

^{2}+ OC

^{2}- OQ

^{2}

**...(vii)**

Adding (iv), (v) and (vi), we get,

AQ

^{2}+CP

^{2}+ BR

^{2}= AO

^{2}- OR

^{2}+ BO

^{2}- OP

^{2}+ OC

^{2}- OQ

^{2}

**...(viii)**

From (

*vii*) and (

*viii*), we get,

AR

^{2}+ BP

^{2}+ CQ

^{2}= AQ

^{2}+ CP

^{2}+ BR

^{2}Hence, proved.

**10. Diagonals of rhombus ABCD intersect each other at point O. Prove that:**

**OA ^{2} + OC^{2} = 2AD^{2}**

**Answer**

Diagonals of the rhombus are perpendicular to each other.

In quadrilateral ABCD, ∠AOD = ∠COD = 90°,

So, Î”AOD and Î”COD are right - angled triangles.

In Î”AOD using Pythagoras theorem,

AD

^{2}= OA

^{2}+ OD

^{2}

⇒ OA

^{2}= AD

^{2}- OD

^{2}

**...(i)**

In Î”COD using Pythagoras theorem,

CD

^{2}= OC

^{2}+ OD

^{2}

⇒ OC

^{2}= CD

^{2}- OD

^{2}

**...(ii)**

LHS = OA

^{2}+ OC

^{2}

= AD

^{2}- OD

^{2}+ CD

^{2}- OD

^{2}

**[from (i) and (ii)]**

= AD

^{2}+ CD

^{2}- 2OD

^{2}

= AD

^{2}+ AD

^{2}- 2(BD/2)

^{2}

**[AD = CD and OD = BD/2]**

= 2AD

^{2}- BD

^{2}/2

= RHS

**11. In the figure AB = BC and AD is perpendicular to CD. Prove that:**

**AC ^{2} = 2BC. DC.**

**Answer**

We consider the Î”ACD and applying Pythagoras theorem we get,

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

= (AB^{2} - DB^{2}) + (DB+BC)^{2}

=BC^{2} - DB^{2} + DB^{2} + BC^{2} + 2DB.BC **(Given, AB = BC)**

= 2BC^{2} + 2DB.BC

= 2BC(BC+DB)

= 2BC.DC

Hence, Proved.

**12. In an isosceles triangle ABC; AB = AC and D is point on BC produced. Prove that:**

**AD ^{2} = AC^{2} + BD.CD.**

**Answer**

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

We consider the rt. angled Î”AED and applying Pythagoras theorem we get,

AD

^{2}= AE

^{2}+ ED

^{2}

⇒ AD

^{2}= AE

^{2}+(EC+CD)

^{2}

**...(i) [∵ ED = EC + CD]**

Similarly, in △AEC,

AC

^{2}= AE

^{2}+ EC

^{2}

⇒ AE

^{2}= AC

^{2}- EC

^{2}

**...(ii)**

putting AE

^{2}= AC

^{2}- EC

^{2}in (i), we get,

AD

^{2}= AC

^{2}- EC

^{2}+ (EC + CD)

^{2}

= AC

^{2}+ CD(CD+2EC)

AD

^{2}= AC

^{2}+ BD.CD

**[∵ 2EC + CD = BD]**

Hence, Proved

**13. In triangle ABC, angle A = 90**

^{o}, CA = AB and D is point on AB produced. Prove that DC^{2}-BD^{2}= 2AB.AD.**Answer**

We consider the rt. angled and applying Pythagoras theorem we get,

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

⇒ CD^{2} = AC^{2} + (AB+ BD)^{2} **[∵ AD = AB + BD]**

⇒ CD^{2} = AC^{2} + AB^{2} + BD^{2} + 2AB. BD **...(i)**

Similarly, in Î”ABC,

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

⇒ BC^{2} = 2AB^{2 } **[AB= AC]**

⇒ AB^{2} = (1/2)BC^{2} **...(ii)**

Putting, AB^{2} from (ii) in (i) we get,

CD^{2} = AC^{2} + (1/2)BC^{2} + BD^{2} + 2AB.BD

⇒ CD^{2} - BD^{2} = AB^{2} + AB^{2} + 2AB.(AD-AB)

⇒ CD^{2} - BD^{2} = AB^{2} + AB^{2} +2AB.AD - 2AB^{2}

⇒ CD^{2} - BD^{2} = 2AB. AD

⇒ DC^{2} - BD^{2} = 2AB.AD

Hence, Proved.

**14. In triangle ABC, =AB = AC and BD is perpendicular to AC. Prove that: BD**

^{2}– CD^{2}= 2CD × AD.**Answer**

In right angled Î”ADB

AB

^{2}= AD

^{2}+ BD

^{2}

⇒ AD

^{2}= AB

^{2}- BD

^{2}

**....(i)**

⇒ AC = AD +DC

⇒ AC

^{2}= (AD + DC)

^{2}

^{2}= AD

^{2}+ DC

^{2}+2AD×DC

⇒ AC

^{2}= AB

^{2}- BD

^{2}+ DC

^{2}+ 2AD×DC

**[from (i)]**

⇒ AC

^{2}= AC

^{2}- BD

^{2}+ DC

^{2}+ 2AD × DC

**[AB = AC]**

⇒ BD

^{2}- DC

^{2}= 2AD×DC