# SELINA Solutions for Class 9 Maths Chapter 10 - Isosceles Triangle

Learn all about isosceles triangles on TopperLearning with Selina Solutions for ICSE Class 9 Mathematics Chapter 10 Isosceles Triangle. Learn about calculating the angle of an isosceles triangle as per the given data. Also, get a grasp of the properties of a triangle and where to use them. You can also understand how to use the properties for finding the measurements asked in a given Maths question.

When you learn and revise with TopperLearning’s Selina textbook solutions, you will understand the methods for proving that a particular triangle is an isosceles triangle. To strengthen your hold on the concept, you can revise the chapter-based problems using our ICSE Class 9 Maths Frank solutions. With practice, your Maths skills will get better and so will your chances of scoring top marks.

## Chapter 10 - Isosceles Triangle Exercise Ex. 10(A)

In the figure alongside,

AB = AC

_{}A = 48^{o} and

_{}ACD = 18^{o}.

Show that BC = CD.

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.

Calculate:

(i) _{}ADC

(ii) _{}ABC

(iii) _{}BAC

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

Proof:

(i)

_{}

_{}

(ii)

_{}

(iii)

_{}

In the following figure, AB = AC; BC = CD and DE is parallel to BC. Calculate:

(i) _{}CDE

(ii) _{}DCE

_{}

_{}

(i)

_{}

(ii)

Calculate x:

(i)

(ii)

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

_{}

_{}

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

_{}

_{}

In the figure, given below, AB = AC. Prove that: _{}BOC = _{}ACD.

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

_{}

Now,

_{}

_{}

In the figure given below, LM = LN; angle PLN = 110^{o}. Calculate:

(i) _{}LMN

(ii) _{}MLN

Given: _{}

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

In quad. PQNL,

_{}

_{}

(ii)

_{}

An isosceles triangle ABC has AC = BC. CD bisects AB at D and _{}CAB = 55^{o}.

Find: (i) _{}DCB (ii) _{}CBD.

_{}

_{}

Now,

_{}

Find x:

Let us name the figure as following:

_{}

_{}

For x:

_{}

In the triangle ABC, BD bisects angle B and is perpendicular to AC. If the lengths of the sides of the triangle are expressed in terms of x and y as shown, find the values of x and y.

_{}

Therefore,

AD=DC

_{}

and AB = BC

_{}

Substituting the value of x from (i)

_{}

Putting y = 3 in (i)

x = 3 + 1

_{}x = 4

In the given figure; AE // BD, AC // ED and AB = AC. Find _{}a, _{}b and _{}c.

Let P and Q be the points as shown below:

Given: _{}

_{}

_{}

_{}

_{}

In the following figure; AC = CD, AD = BD and _{}C = 58^{o}.

Find angle CAB.

_{}

Now,

_{}

_{}

In the figure of q. no. 11 given above, if AC = AD = CD = BD; find angle ABC.

_{}

_{}

In triangle ABC; AB = AC and _{}A : _{}B = 8 : 5; find angle A.

Let _{}

Given: AB = AC

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

_{}

In triangle ABC; _{}A = 60^{o}, _{}C = 40^{o}, and bisector of angle ABC meets side AC at point P. Show that BP = CP.

_{}

Now,

BP is the bisector of _{}

_{}

_{}

In triangle ABC; angle ABC = 90^{o} and P is a point on AC such that _{}PBC = _{}PCB. Show that: PA = PB.

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]

ABC is an equilateral triangle. Its side BC is produced upto point E such that C is mid-point of BE. Calculate the measure of angles ACE and AEC.

In triangle ABC, D is a point in AB such that AC = CD = DB. If ∠B = 28°, find the angle ACD.

In the given figure, AD = AB = AC, BD is parallel to CA and angle ACB = 65°. Find angle DAC.

Prove that a triangle ABC is isosceles, if:

(i) altitude AD bisects angles BAC, or

(ii) bisector of angle BAC is perpendicular to base BC.

(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.

In the given figure; AB = BC and AD = EC.

Prove that: BD = BE.

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)

## Chapter 10 - Isosceles Triangle Exercise Ex. 10(B)

If the equal sides of an isosceles triangle are produced, prove that the exterior angles so formed are obtuse and equal.

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.

In the given figure, AB = AC. Prove that:

(i) DP = DQ

(ii) AP = AQ

(iii) AD bisects angle A

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.

In triangle ABC, AB = AC; BE _{}AC and CF _{}AB. Prove that:

(i) BE = CF

(ii) AF = AE

(i)

_{}

(ii)Since _{}

_{}

In isosceles triangle ABC, AB = AC. The side BA is produced to D such that BA = AD. Prove that: _{}BCD = 90^{o}

Const: Join CD.

_{}

_{}

Adding (i) and (ii)

_{}

_{}

(i) In triangle ABC, AB = AC and _{}= 36°. If the internal bisector of _{}meets AB at point D, prove that AD = BC.

(ii) If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles.

_{}

_{ }

_{ }

_{ }

Prove that the bisectors of the base angles of an isosceles triangle are equal.

_{}

_{}

In the given figure, AB = AC and _{}DBC = _{}ECB = 90^{o}

Prove that:

(i) BD = CE

(ii) AD = AE

_{}

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

DBC = ECB …….(ii)

Subtracting (i) from (ii)

_{}

ABC and DBC are two isosceles triangles on the same side of BC. Prove that:

(i) DA (or AD) produced bisects BC at right angle.

(ii) _{}BDA = _{}CDA.

DA is produced to meet BC in L.

_{}

_{}

Subtracting (i) from (ii)

_{}

_{}

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

_{}

_{}

From (vi) and (vii)

_{}

Now,

_{}

The bisectors of the equal angles B and C of an isosceles triangle ABC meet at O. Prove that AO bisects angle A.

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.

Prove that the medians corresponding to equal sides of an isosceles triangle are equal.

_{}

Use the given figure to prove that, AB = AC.

_{}

_{}

_{}

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

_{}

_{}

In the given figure; AE bisects exterior angle CAD and AE is parallel to BC.

Prove that: AB = AC.

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]

In an equilateral triangle ABC; points P, Q and R are taken on the sides AB, BC and CA respectively such that AP = BQ = CR. Prove that triangle PQR is equilateral.

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.

In triangle ABC, altitudes BE and CF are equal. Prove that the triangle is isosceles.

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.

Through any point in the bisector of angle, a straight line is drawn parallel to either arm of the angle. Prove that the triangle so formed is isosceles.

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.

In triangle ABC; AB = AC. P, Q and R are mid-points of sides AB, AC and BC respectively. Prove that:

(i) PR = QR(ii) BQ = CP

(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

From the following figure, prove that:

(i) _{}ACD = _{}CBE

(ii) AD = CE

(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)

Equal sides AB and AC of an isosceles triangle ABC are produced. The bisectors of the exterior angle so formed meet at D. Prove that AD bisects angle A.

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.

ABC is a triangle. The bisector of the angle BCA meets AB in X. A point Y lies on CX such that AX = AY.

Prove that _{}CAY = _{}ABC.

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

In the following figure; IA and IB are bisectors of angles CAB and CBA respectively. CP is parallel to IA and CQ is parallel to IB.

Prove that:

PQ = The perimeter of the _{}ABC.

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

Sides AB and AC of a triangle ABC are equal. BC is produced through C upto a point D such that AC = CD. D and A are joined and produced upto point E. If angle BAE = 108^{o}; find angle ADB.

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}

The given figure shows an equilateral triangle ABC with each side 15 cm. Also, DE//BC, DF//AC and EG//AB.

If DE + DF + EG = 20 cm, find FG.

_{}

_{ }

_{ }

_{ }

_{ }

If all the three altitudes of a triangle are equal, the triangle is equilateral. Prove it.

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

BE = CF[Given]

BC = BC[Common]

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

Therefore, ABC is an equilateral triangle.

In a _{}ABC, the internal bisector of angle A meets opposite side BC at point D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Show that _{}ACE is isosceles.

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.

In triangle ABC, bisector of angle BAC meets opposite side BC at point D. If BD = CD, prove that _{}ABC is isosceles.

Produce AD upto E such that AD = DE.

Hence, ABC is an isosceles triangle.

In _{}ABC, D is point on BC such that AB = AD = BD = DC. Show that:

_{}ADC: _{}C = 4: 1.

Since AB = AD = BD

_{}is an equilateral triangle.

_{}

Again in_{}

AD = DC

_{}

Using the information given in each of the following figures, find the values of a and b.

[Given: CE = AC]

_{(i)}

_{}

(ii)

_{}

_{ }

_{ }

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change