# Class 9 SELINA Solutions Maths Chapter 10 - Isosceles Triangle

## Isosceles Triangle Exercise Ex. 10(A)

### Solution 1(a)

Correct option: (i) AB = AC

In ∆BAD and ∆CAD

∠B = ∠C

∠BAD = ∠CAD

AD is common

Hence,

∆BAD ≅ ∆CAD … (A.A.S.)

∴ AB = AC … (c.p.c.t.)

### Solution 1(b)

Correct option: (ii) BD = CD

In ∆BAD and ∆CAD

AB = AC … given

AD common

∠ADB = ∠ADC … (each 90^{o})

∆BAD ≅ ∆CAD … (R.H.S.)

∴ BD = CD … (c.p.c.t.)

### Solution 1(c)

Correct option: (iv) 28°

∆ADB is isosceles triangle

∴ ∠ABD = ∠BAD = 65^{o}

Now,

∠ADC = ∠ABD + ∠BAD …(external angle theorem)

∴ ∠ADC = 130^{o}

In ∆ADC,

∠ADC + ∠DAC + ∠ACD = 180 … (sum of angles in a triangle)

130 + 22 + ∠ACD = 180

∠ACD = 28^{o}

### Solution 1(d)

Correct option: (iii) AD = AE

∆ABC is isosceles triangle

∴ ∠ABD = ∠ACE

Now, in ∆ABD and ∆ACE

AB = AC

∠ABD = ∠ACE

Now,

BE = DC

BD + DE = DE + EC

BD = EC

∆ABD ≅ ∆ACE … (S.A.S.)

### Solution 1(e)

Correct option: (i) isosceles but not congruent

In ΔABC and ΔPQR,

AB = AC,

∴ ΔABC is isosceles

∠C = ∠B …(1)

Also,

∠C = ∠P …(2)

∠B = ∠Q …(3)

∴∠P = ∠Q … (from (1), (2) and (3))

∴ ΔPQR is isosceles

But, we can't prove ΔABC ≅ ΔPQR using the given data.

Therefore, triangles are isosceles but not congruent.

### Solution 2

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 3

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

Proof:

(i)

_{}

_{}

(ii)

_{}

(iii)

_{}

### Solution 4

_{}

_{}

(i)

_{}

(ii)

### Solution 5

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

_{}

_{}

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

_{}

_{}

### Solution 6

Let _{}ABO =_{}OBC = x and _{}ACO = _{}OCB = y

_{}

Now,

_{}

_{}

### Solution 7

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 8

_{}

_{}

Now,

_{}

### Solution 9

Let us name the figure as following:

_{}

_{}

For x:

_{}

### Solution 10

_{}

Therefore,

AD=DC

_{}

and AB = BC

_{}

Substituting the value of x from (i)

_{}

Putting y = 3 in (i)

x = 3 + 1

_{}x = 4

### Solution 11

Let P and Q be the points as shown below:

Given: _{}

_{}

_{}

_{}

_{}

### Solution 12

_{}

Now,

_{}

_{}

### Solution 13

_{}

_{}

### Solution 14

Let _{}

Given: AB = AC

_{}[Angles opp. to equal sides are equal]

_{}

### Solution 15

_{}

Now,

BP is the bisector of _{}

_{}

_{}

## Isosceles Triangle Exercise Ex. 10(B)

### Solution 1(a)

Correct option: (ii) ΔABD ≅ ΔFEC

In ∆ABD and ∆FEC

AB = EF

BC =DE

∴ BC + CD = DE + CD

∴ BD = EC

∠B = ∠E … (both 90^{o})

∆ABD ≅ ∆FEC … (S.A.S.)

### Solution 1(b)

Correct option: (iii) PQ = PR

In ∆PQO and ∆PRO

∠PQO = ∠PRO … (each
90^{o})

∠POQ = ∠POR … given

PO is common

∆PQO ≅ ∆PRO … (A.A.S.)

PQ = PR … (c.p.c.t.)

### Solution 1(c)

Correct option: (iv) x = 16, y = 8

∠A = ∠C

∴ ∆ABC is isosceles

∴ AB = BC

∴ 2x = 3y+8 … (1)

Also, in ∆ABD and ∆CBD

∠ABD = ∠CBD … given

AB = BC … given

BD is common

∆ ABD ≅ ∆ CBD … (S.A.S.)

∴ AD = CD … (c.p.c.t.)

∴ x = 2y …(2)

From (1) and (2)

4y = 3y + 8

y = 8

x = 16

### Solution 1(d)

Correct option: (iv) ∆ABX ≅ ∆BAY

In ∆ABX and ∆BAY

∠XAB = ∠YBA … (each
90^{o})

AX = BY … given

AB is common

∆ ABX ≅ ∆BAY … (S.A.S.)

### Solution 1(e)

Correct option: (ii) ∠PBC = ∠PCB

Assuming that quadrilateral ABCD is a Square

We have

DC = AB

∠CDP = ∠BAP … (each
90^{o})

DP = AP … (P is mid-point of side AD)

Hence,

In ∆ CDP and ∆BAP

We can say that

∆ CDP ≅ ∆BAP … (S.A.S.)

CP = BP … (c.p.c.t.)

Now, in ∆BPC

We have

CP = BP

Hence, ∆BPC is isosceles

∴ ∠PBC = ∠PCB

### Solution 2

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 3

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 4

(i)

_{}

(ii)Since _{}

_{}

### Solution 5

Const: Join CD.

_{}

_{}

Adding (i) and (ii)

_{}

_{}

### Solution 6(i)

### Solution 6(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

_{}

_{}

### Solution 8

_{}

_{}DBC = _{}ECB = 90^{o}[Given]

DBC = ECB …….(ii)

Subtracting (i) from (ii)

_{}

### Solution 9

DA is produced to meet BC in L.

_{}

_{}

Subtracting (i) from (ii)

_{}

_{}

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

_{}

_{}

From (vi) and (vii)

_{}

Now,

_{}

### Solution 10

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

_{}

### Solution 12

_{}

_{}

_{}

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

_{}

_{}

### Solution 13

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 14

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 15

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 16

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 17

(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 18

(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 19

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 20

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

## Isosceles Triangle Exercise Test Yourself

### Solution 1

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 2

### Solution 3

### Solution 4

### Solution 5

(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 6

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)

### Solution 7

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 8

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 9

_{}

### Solution 10

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

BE = CF[Given]

BC = BC[Common]

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

Therefore, ABC is an equilateral triangle.

### Solution 11

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 12

Produce AD upto E such that AD = DE.

Hence, ABC is an isosceles triangle.

### Solution 13

Since AB = AD = BD

_{}is an equilateral triangle.

_{}

Again in_{}

AD = DC

_{}

### Solution 14

_{(i)}

_{}

(ii)