**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

Â½ B = Â½ C

â‡’ OBC = OCB (Angle bisectors)

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

(ii) In Î”AOB and Î”AOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, Î”AOB Î”AOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects A.

**2. In Î”ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that Î”ABC 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 Î”ADB and Î”ADC,

AD = AD (It is the Common arm)

ADB = ADC

BD = CD (Since AD is the perpendicular bisector)

So, Î”ADB Î”ADC byÂ **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 Î”AEB and Î”AFC 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)

âˆ´ Î”AEB Î”AFC 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) Î”ABE Î”ACF

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

**Solution:**

It is given that BE = CF

(i) In Î”ABE and Î”ACF,

A = A (It is the common angle)

AEB = AFC (They are right angles)

BE = CF (Given in the question)

âˆ´ Î”ABE Î”ACF byÂ **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 Î”ABD and Î”ACD 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, Î”ABD Î”ACD.

âˆ´ ABD = ACD by CPCT.

**6. Î”ABC 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 Î”ABC,

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 Î”ACD,

AD = AB

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

Now,

In Î”ABC,

CAB + ACB + ABC = 180Â°

So, CAB + 2ACB = 180Â°

â‡’ CAB = 180Â° â€“ 2ACB â€” (i)

Similarly, in Î”ADC,

CAD = 180Â° â€“ 2ACD â€” (ii)

also,

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

Adding (i) and (ii) we get,

CAB + CAD = 180Â° â€“ 2ACB+180Â° â€“ 2ACD

â‡’ 180Â° = 360Â° â€“ 2ACB-2ACD

â‡’ 2(ACB+ACD) = 180Â°

â‡’ BCD = 90Â°

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

**Solution:**

In the question, it is given that

A = 90Â° and AB = AC

AB = AC

â‡’ B = C (They are angles opposite to the equal sides and so, they are equal)

Now,

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

âˆ´ 90Â° + 2B = 180Â°

â‡’ 2B = 90Â°

â‡’ B = 45Â°

So, B = C = 45Â°

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

**Solution:**

Let ABC be an equilateral triangle as shown below:

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

â‡’ A = B =C (Sides opposite to the equal angles are equal.)

Also, we know that

A+B+C = 180Â°

â‡’ 3A = 180Â°

â‡’ A = 60Â°

âˆ´ A = B = C = 60Â°

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