Question

∆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.

Answer 1

In ∆ABC, 

AB = AC (Given) 

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

In ∆ACD, 

AC = AD 

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

In ∆BCD, 

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

∠ACB + ∠ACB + ∠ACD + ∠ACD = 180º 

2( ∠ACB +∠ ACD) = 180º 

2( ∠BCD) = 180º 

∠BCD = 90º

Answer 2

Given: In an isosceles triangle \(\triangle ABC\), AB = AC.
To prove: \(\angle BCD = 90^\circ\).

Proof:
1. Since AB = AC, the angles opposite to these equal sides are also equal. Thus, \(\angle ACB = \angle ABC\). Let these angles be x.
\(\angle ACB = \angle ABC = x \quad \text{(1)}\)

2. In \(\triangle ACD\), since AC = AD (because AB = AD), the angles opposite to these equal sides are also equal. Thus, \(\angle ADC = \angle ACD\). Let these angles be y.
\(\angle ADC = \angle ACD = y \quad \text{(2)}\)

3. Now, consider \(\triangle BCD\). The exterior angle \(\angle BCD\) can be expressed as the sum of \(\angle ACB\) and \(\angle ACD\) because they are adjacent to \(\angle BCD\).
\(\angle BCD = \angle ACB + \angle ACD = x + y \quad \text{(3)}\)

4. Using the angle sum property of triangles in \(\triangle BCD\):
\(\angle ABC + \angle BCD + \angle ADC = 180^\circ\)

5. Substitute the values from equations (1), (2), and (3):
\(x + (x + y) + y = 180^\circ\)  Simplify the equation:
\(2x + 2y = 180^\circ\)
\(2(x + y) = 180^\circ\)
Divide both sides by 2:
\(x + y = 90^\circ\)

6. Therefore, from equation (3):
\(\angle BCD = x + y = 90^\circ\)

Hence, we have proved that \(\angle BCD = 90^\circ\).