Triangles | Exercise 7.2 | NCERT Solution Maths 9 | Chapter 7

 Exercise: 7.2 (Page No: 123)

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

Solution:

Given:

AB = AC and

the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

∠B = ∠C

½ ∠B = ½ ∠C

⇒ ∠OBC = ∠OCB (Angle bisectors)

∴ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, ΔAOB ≅ ΔAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects ∠A.

2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.

Solution:

It is given that AD is the perpendicular bisector of BC

To prove:

AB = AC

Proof:

In ΔADB and ΔADC,

AD = AD (It is the Common arm)

∠ADB = ∠ADC

BD = CD (Since AD is the perpendicular bisector)

So, ΔADB ≅ ΔADC by SAS congruency criterion.

Thus,

AB = AC (by CPCT)

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Solution:

Given:

(i) BE and CF are altitudes.

(ii) AC = AB

To prove:

BE = CF

Proof:

Triangles ΔAEB and ΔAFC are similar by AAS congruency since

∠A = ∠A (It is the common arm)

∠AEB = ∠AFC (They are right angles)

AB = AC (Given in the question)

∴ ΔAEB ≅ ΔAFC and so, BE = CF (by CPCT).

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that

(i) ΔABE ≅ ΔACF

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

Solution:

It is given that BE = CF

(i) In ΔABE and ΔACF,

∠A = ∠A (It is the common angle)

∠AEB = ∠AFC (They are right angles)

BE = CF (Given in the question)

∴ ΔABE ≅ ΔACF by AAS congruency condition.

(ii) AB = AC by CPCT and so, ABC is an isosceles triangle.

5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

Solution:

In the question, it is given that ABC and DBC are two isosceles triangles.

We will have to show that ∠ABD = ∠ACD

Proof:

Triangles ΔABD and ΔACD are similar by SSS congruency since

AD = AD (It is the common arm)

AB = AC (Since ABC is an isosceles triangle)

BD = CD (Since BCD is an isosceles triangle)

So, ΔABD ≅ ΔACD.

∴ ∠ABD = ∠ACD by CPCT.

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

Solution:

It is given that AB = AC and AD = AB

We will have to now prove ∠BCD is a right angle.

Proof:

Consider ΔABC,

AB = AC (It is given in the question)

Also, ∠ACB = ∠ABC (They are angles opposite to the equal sides and so, they are equal)

Now, consider ΔACD,

AD = AB

Also, ∠ADC = ∠ACD (They are angles opposite to the equal sides and so, they are equal)

Now,

In ΔABC,

∠CAB + ∠ACB + ∠ABC = 180°

So, ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — (i)

Similarly, in ΔADC,

∠CAD = 180° – 2∠ACD — (ii)

also,

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii) we get,

∠CAB + ∠CAD = 180° – 2∠ACB+180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB-2∠ACD

⇒ 2(∠ACB+∠ACD) = 180°

⇒ ∠BCD = 90°

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

Solution:

In the question, it is given that

∠A = 90° and AB = AC

AB = AC

⇒ ∠B = ∠C (They are angles opposite to the equal sides and so, they are equal)

Now,

∠A+∠B+∠C = 180° (Since the sum of the interior angles of the triangle)

∴ 90° + 2∠B = 180°

⇒ 2∠B = 90°

⇒ ∠B = 45°

So, ∠B = ∠C = 45°

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

Solution:

Let ABC be an equilateral triangle as shown below:

Here, BC = AC = AB (Since the length of all sides is same)

⇒ ∠A = ∠B =∠C (Sides opposite to the equal angles are equal.)

Also, we know that

∠A+∠B+∠C = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

∴ ∠A = ∠B = ∠C = 60°

So, the angles of an equilateral triangle are always 60° each.