# Class 9 SELINA Solutions Maths Chapter 10 - Isosceles Triangle

## Isosceles Triangle Exercise Ex. 10(A)

### Solution 1

In _{}

_{}BAC + _{}ACB + _{}ABC = 180^{0}

48^{0} + _{}ACB + _{}ABC = 180^{0}

But _{}ACB = _{}ABC [AB = AC]

2_{}ABC = 180^{0} - 48^{0}

2_{}ABC = 132^{0}

_{}ABC = 66^{0} = _{}ACB ……(i)

_{}ACB = 66^{0}

_{}ACD + _{}DCB = 66^{0}

18^{0} + _{}DCB = 66^{0}

_{}DCB = 48^{0} ………(ii)

Now, In _{}

_{}DBC = 66^{0} [From (i), Since _{}ABC = _{}DBC]

_{}DCB = 48^{0 } [From (ii)]

_{}BDC = 180^{0} - 48^{0} - 66^{0}

_{}BDC = 66^{0}

Since _{}BDC = _{}DBC

Therefore, BC = CD

Equal angles have equal sides opposite to them.

### Solution 2

Given: _{}ACE = 130^{0}; AD = BD = CD

Proof:

(i)

_{}

_{}

(ii)

_{}

(iii)

_{}

### Solution 3

_{}

_{}

(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 360^{o}.

In quad. PQNL,

_{}

_{}

(ii)

_{}

### Solution 7

_{}

_{}

Now,

_{}

### Solution 8

Let us name the figure as following:

_{}

_{}

For x:

_{}

### Solution 9

_{}

Therefore,

AD=DC

_{}

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: _{}

_{}

_{}

_{}

_{}

### Solution 11

_{}

Now,

_{}

_{}

### Solution 12

_{}

_{}

### 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 16

### Solution 17

### Solution 18

### Solution 19

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

Then we have to prove that the ΔABC is isosceles.

In triangles ADB and ADC,

∠BAD = ∠CAD (AD is bisector of ∠BAC)

AD = AD (common)

∠ADB = ∠ADC (Each equal to 90°)

⇒ ΔADB ≅ ΔADC (by ASA congruence criterion)

⇒ 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.

In triangles ADB and ADC,

∠BAD = ∠CAD (AD is bisector of ∠BAC)

AD = AD (common)

∠ADB = ∠ADC (Each equal to 90°)

⇒ ΔADB ≅ ΔADC (by ASA congruence criterion)

⇒ 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

Const: Join AD.

_{}

(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.

### Solution 3

(i)

_{}

(ii)Since _{}

_{}

### Solution 4

Const: Join CD.

_{}

_{}

Adding (i) and (ii)

_{}

_{}

### Solution 5(i)

### 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 6

_{}

_{}

### Solution 7

_{}

_{}DBC = _{}ECB = 90^{o}[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 10

_{}

### 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 = 90^{0}[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)

_{}ADE = _{}DAE

_{}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 = 180^{0} …….(ii)[DCB is a straight line]

_{}ABC + _{}CBE = 180^{0} ……..(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 = 180^{0} - _{}B[ABE is a straight line]

_{}[BD is bisector of _{}CBE]

_{}

Similarly,

_{}BCF = 180^{0} - _{}C[ACF is a straight line]

_{}[CD is bisector of _{}BCF]

_{}

Now,

_{}

In _{}BCD,

_{}

_{}BD = CD

In _{}ABD and _{}ACD,

AB = AC[Given]

AD = AD[Common]

BD = CD[Proved]

_{}

Therefore, AD bisects_{}A.

### 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

_{}108^{0} = 3 + ADB

But AB = AC

_{}3 = _{}2

108^{0} = 2 + ADB ……(i)

Now,

In ACD,

2=1+ ADB

But AC = CD

1 = ADB

2 = ADB + ADB

2 = 2ADB

Putting this value in (i)

**108**^{0} = 2ADB + ADB

**3ADB = 108**^{0}

**ADB = 36**^{0}

### Solution 22

_{}

_{ }

_{ }

_{ }

_{ }

### Solution 23

In right _{}BEC and _{}BFC,

BE = CF[Given]

BC = BC[Common]

_{}BEC = _{}BFC[each = 90^{0}]

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_{}

AD = DC

_{}

### Solution 27

_{(i)}

_{}

(ii)

_{}

_{ }

_{ }