Question

In the adjoining figure, ABC is an isosceles triangle in which AB = AC. Its side BA is produced to D such that AD = AB. Show that ∠BCD = 90°.

Answer 1

Given:

• Triangle ABC with AB = AC.
• BA is produced to D, so B,A,D are collinear and AD = AB.

To Prove: ∠BCD = 90°.

Proof (Step-wise):

1. From AB = AC (given) and AD = AB (given) we get AB = AC = AD, so points B, C and D are all at the same distance from A.

2. Hence, B, C and D lie on a circle with center A (radius = AB). All three points are equidistant from A.

3. Because A lies on BD and AB = AD, A is the midpoint of BD, so BD passes through the center A, therefore BD is a diameter of that circle.

4. By the theorem that an angle subtended by a diameter is a right angle (angle in a semicircle), the angle subtended at any point on the circle by the diameter BD is 90°. Thus ∠BCD = 90°.

∠BCD = 90°, as required.