# NCERT solution class 9 chapter 7 Triangles exercise 7.2 mathematics

## EXERCISE 7.2

#### 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

(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.

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

∴AB = AC (By CPCT)

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

#### 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.

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.

(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.

In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

∴ 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.

In ΔABC,

AB = AC (Given)

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

In ΔACD,

⇒ ∠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.

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.

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.