## EXERCISE 7.2

#### Page No 123:

#### Question 1:

In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

(i) OB = OC (ii) AO bisects ∠A

#### Answer:

(i) It is given that in triangle ABC, AB = AC

⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are equal)

⇒ ∠ACB = ∠ABC

⇒ ∠OCB = ∠OBC

⇒ OB = OC (Sides opposite to equal angles of a triangle are also equal)

(ii) In ΔOAB and ΔOAC,

AO =AO (Common)

AB = AC (Given)

OB = OC (Proved above)

Therefore, ΔOAB ≅ ΔOAC (By SSS congruence rule)

⇒ ∠BAO = ∠CAO (CPCT)

⇒ AO bisects ∠A.

#### Question 2:

In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.

#### Answer:

In ΔADC and ΔADB,

AD = AD (Common)

∠ADC =∠ADB (Each 90º)

CD = BD (AD is the perpendicular bisector of BC)

∴ ΔADC ≅ ΔADB (By SAS congruence rule)

∴AB = AC (By CPCT)

Therefore, ABC is an isosceles triangle in which AB = AC.

#### Page No 124:

#### Question 3:

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see the given figure). Show that these altitudes are equal.

#### Answer:

In ΔAEB and ΔAFC,

∠AEB and ∠AFC (Each 90º)

∠A = ∠A (Common angle)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)

⇒ BE = CF (By CPCT)

#### Question 4:

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that

(i) ABE ≅ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

#### Answer:

(i) In ΔABE and ΔACF,

∠AEB = ∠AFC (Each 90º)

∠A = ∠A (Common angle)

BE = CF (Given)

∴ ΔABE ≅ ΔACF (By AAS congruence rule)

(ii) It has already been proved that

ΔABE ≅ ΔACF

⇒ AB = AC (By CPCT)

#### Question 5:

ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that ∠ABD = ∠ACD.

#### Answer:

Let us join AD.

In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common side)

∴ ΔABD ΔACD (By SSS congruence rule)

⇒ ∠ABD = ∠ACD (By CPCT)

#### Question 6:

ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.

#### Answer:

In ΔABC,

AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)

In ΔACD,

AC = AD

⇒ ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)

In ΔBCD,

∠ABC + ∠BCD + ∠ADC = 180º (Angle sum property of a triangle)

⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º

⇒ 2(∠ACB + ∠ACD) = 180º

⇒ 2(∠BCD) = 180º

⇒ ∠BCD = 90º

#### Question 7:

ABC is a right angled triangle in which ∠A = 90º and AB = AC. Find ∠B and ∠C.

#### Answer:

It is given that

AB = AC

⇒ ∠C = ∠B (Angles opposite to equal sides are also equal)

In ΔABC,

∠A + ∠B + ∠C = 180º (Angle sum property of a triangle)

⇒ 90º + ∠B + ∠C = 180º

⇒ 90º + ∠B + ∠B = 180º

⇒ 2 ∠B = 90º

⇒ ∠B = 45º

∴ ∠B = ∠C = 45º

#### Question 8:

Show that the angles of an equilateral triangle are 60º each.

#### Answer:

Let us consider that ABC is an equilateral triangle.

Therefore, AB = BC = AC

AB = AC

⇒ ∠C = ∠B (Angles opposite to equal sides of a triangle are equal)

Also,

AC = BC

⇒ ∠B = ∠A (Angles opposite to equal sides of a triangle are equal)

Therefore, we obtain

∠A = ∠B = ∠C

In ΔABC,

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠A + ∠A = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

⇒ ∠A = ∠B = ∠C = 60°

Hence, in an equilateral triangle, all interior angles are of measure 60º.