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# Class 9 SELINA Solutions Maths Chapter 10 - Isosceles Triangle

## Isosceles Triangle Exercise Ex. 10(A)

### Solution 1

In

BAC + ACB + ABC = 1800

480 + ACB + ABC = 1800

But ACB = ABC [AB = AC]

2ABC = 1800 - 480

2ABC = 1320

ABC = 660 = ACB ……(i)

ACB = 660

ACD + DCB = 660

180 + DCB = 660

DCB = 480 ………(ii)

Now, In

DBC = 660 [From (i), Since ABC = DBC]

DCB = 480 [From (ii)]

BDC = 1800 - 480 - 660

BDC = 660

Since BDC = DBC

Therefore, BC = CD

Equal angles have equal sides opposite to them.

### Solution 2

Given: ACE = 1300; AD = BD = CD

Proof:

(i)

(ii)

(iii)

(i)

(ii)

### Solution 4

(i) Let the triangle be ABC and the altitude be AD.

(ii) Let triangle be ABC and altitude be AD.

### Solution 5

Let ABO =OBC = x and ACO = OCB = y

Now,

### Solution 6

Given:

(i) We know that the sum of the measure of all the angles of a quadrilateral is 360o.

(ii)

Now,

### Solution 8

Let us name the figure as following:

For x:

### Solution 9

Therefore,

and AB = BC

Substituting the value of x from (i)

Putting y = 3 in (i)

x = 3 + 1

x = 4

### Solution 10

Let P and Q be the points as shown below:

Given:

Now,

### Solution 13

Let

Given: AB = AC

[Angles opp. to equal sides are equal]

### Solution 14

Now,

BP is the bisector of

### Solution 15

Let PBC = PCB = x

In the right angled triangle ABC,

and

Therefore in the triangle ABP;

Hence,

PA = PB [sides opp. to equal angles are equal]

### Solution 19

(i)  In ΔABC, let the altitude AD bisects BAC.

Then we have to prove that the ΔABC is isosceles.

ADB = ADC (Each equal to 90°)

AB = AC (cpct)

Hence, ΔABC is an isosceles.

(ii)  In Δ ABC, the bisector of BAC is perpendicular to the base BC. We have to prove that the ΔABC is isosceles.

ADB = ADC (Each equal to 90°)

AB = AC (cpct)

Hence, ΔABC is an isosceles.

### Solution 20

In ΔABC,

AB = BC (given)

BCA = BAC (Angles opposite to equal sides are equal)

BCD = BAE ….(i)

Given, AD = EC

AD + DE = EC + DE (Adding DE on both sides)

AE = CD ….(ii)

Now, in triangles ABE and CBD,

AB = BC (given)

BAE = BCD [From (i)]

AE = CD [From (ii)]

ΔABE ΔCBD

BE = BD (cpct)

## Isosceles Triangle Exercise Ex. 10(B)

### Solution 1

Const: AB is produced to D and AC is produced to E so that exterior angles and is formed.

Since angle B and angle C are acute they cannot be right angles or obtuse angles.

Now,

Therefore, exterior angles formed are obtuse and equal.

### Solution 2

(i)

(ii) We have already proved that

Therefore,BP = CQ[cpct]

Now,

AB = AC[Given]

AB - BP = AC - CQ

AP = AQ

(iii)

Hence, AD bisects angle A.

(i)

(ii)Since

### Solution 4

Const: Join CD.

Adding (i) and (ii)

### Solution 5(ii)

Given:

AD is the angle bisector of ∠BAC, hence ∠BAD = ∠DAC

Also, AD bisects BC, hence BD = DC.

To prove: ∆ABC is isosceles, i.e. AB = AC

Proof:

In ∆ABC

BD = DC … (given)

Now by angle bisector theorem

### Solution 7

DBC = ECB = 90o[Given]

DBC = ECB …….(ii)

Subtracting (i) from (ii)

### Solution 8

DA is produced to meet BC in L.

Subtracting (i) from (ii)

From (iii), (iv) and (v)

From (vi) and (vii)

Now,

### Solution 9

In ABC, we have AB = AC

B = C [angles opposite to equal sides are equal]

Now,

In ABO and ACO,

AB = AC [Given]

OBC = OCB [From (i)]

OB = OC [From (ii)]

Therefore, AO bisects BAC.

### Solution 11

From (i), (ii) and (iii)

### Solution 12

Since AE || BC and DAB is the transversal

Since AE || BC and AC is the transversal

But AE bisects

AB = AC[Sides opposite to equal angles are equal]

### Solution 13

AB = BC = CA…….(i) [Given]

AP = BQ = CR…….(ii) [Given]

Subtracting (ii) from (i)

AB - AP = BC - BQ = CA - CR

BP = CQ = AR …………(iii)

……..(iv) [angles opp. to equal sides are equal]

From (v) and (vi)

PQ = QR = PR

Therefore, PQR is an equilateral triangle.

### Solution 14

In ABE and ACF,

A = A[Common]

AEB = AFC = 900[Given: BE AC; CF AB]

BE = CF[Given]

Therefore, ABC is an isosceles triangle.

### Solution 15

AL is bisector of angle A. Let D is any point on AL. From D, a straight line DE is drawn parallel to AC.

DE || AC [Given]

ADE = DAC….(i) [Alternate angles]

DAC = DAE…….(ii) [AL is bisector of A]

From (i) and (ii)

AE = ED [Sides opposite to equal angles are equal]

Therefore, AED is an isosceles triangle.

### Solution 16

(i)

In ABC,

AB = AC

AP = AQ …….(i)[ Since P and Q are mid - points]

In BCA,

PR = [PR is line joining the mid - points of AB and BC]

PR = AQ……..(ii)

In CAB,

QR = [QR is line joining the mid - points of AC and BC]

QR = AP……(iii)

From (i), (ii) and (iii)

PR = QR

(ii)

AB = AC

B = C

Also,

In BPC and CQB,

BP = CQ

B = C

BC = BC

Therefore, ΔBPCΔCQB   [SAS]

BP = CP

### Solution 17

(i) In ACB,

AC = AC[Given]

ABC = ACB …….(i)[angles opposite to equal sides are equal]

ACD + ACB = 1800 …….(ii)[DCB is a straight line]

ABC + CBE = 1800 ……..(iii)[ABE is a straight line]

Equating (ii) and (iii)

ACD + ACB = ABC + CBE

ACD + ACB = ACB + CBE[From (i)]

ACD = CBE

(ii)

### Solution 18

AB is produced to E and AC is produced to F. BD is bisector of angle CBE and CD is bisector of angle BCF. BD and CD meet at D.

In ABC,

AB = AC[Given]

C = B[angles opposite to equal sides are equal]

CBE = 1800 - B[ABE is a straight line]

[BD is bisector of CBE]

Similarly,

BCF = 1800 - C[ACF is a straight line]

[CD is bisector of BCF]

Now,

In BCD,

BD = CD

In ABD and ACD,

AB = AC[Given]

BD = CD[Proved]

### Solution 19

In ABC,

CX is the angle bisector of C

ACY = BCX ....... (i)

In AXY,

AX = AY [Given]

AXY = AYX …….(ii) [angles opposite to equal sides are equal]

Now XYC = AXB = 180° [straight line]

AYX + AYC = AXY + BXY

AYC = BXY ........ (iii) [From (ii)]

In AYC and BXC

AYC + ACY + CAY = BXC + BCX + XBC = 180°

CAY = XBC [From (i) and (iii)]

CAY = ABC

### Solution 20

Since IA || CP and CA is a transversal

CAI = PCA [Alternate angles]

Also, IA || CP and AP is a transversal

IAB = APC [Corresponding angles]

But CAI = IAB [Given]

PCA = APC

AC = AP

Similarly,

BC = BQ

Now,

PQ = AP + AB + BQ

= AC + AB + BC

= Perimeter of ABC

### Solution 21

In ABD,

BAE = 3 + ADB

1080 = 3 + ADB

But AB = AC

3 = 2

1080 = 2 + ADB ……(i)

Now,

In ACD,

But AC = CD

Putting this value in (i)

### Solution 23

In right BEC and BFC,

BE = CF[Given]

BC = BC[Common]

BEC = BFC[each = 900]

Therefore, ABC is an equilateral triangle.

### Solution 24

DA || CE[Given]

[Corresponding angles]

[Alternate angles]

But [ AD is the bisector of A]

From (i), (ii) and (iii)

AC = AE

ACE is an isosceles triangle.

### Solution 25

Produce AD upto E such that AD = DE.

Hence, ABC is an isosceles triangle.

### Solution 26

Since AB = AD = BD

is an equilateral triangle.

Again in

(i)

(ii)