Question

Let ABC be a triangle such that ∠B = 70° and ∠C = 40°. Suppose D is a point on BC such that AB = AD. Prove that AB > CD.

Answer 1

Given: AB = AD, ∠B = 70° and ∠C = 40°

To prove: AB > CD

Proof:

In ΔABD:

AB = AD (Given)

∠B = ∠ADB (Angles opposite to equal sides are equal)

∠ADB = 70°

∠ADB + ∠ADC = 180° (Linear angles)

∠ADC = 180° − 70°

∠ADC = 110°

In ΔADC:

∠DAC + ∠ADC + ∠ACD = 180° (Sum of the angles of a triangle)

∠DAC + 110° + 40° = 180°

∠DAC = 180° − 150°

∠DAC = 30°

We know that in a triangle, the side opposite to the smallest angle is shortest and the side opposite to the largest angle is longest.

In ΔADC:

∠ACD = 40° > 30° = ∠DAC

⇒ AD > CD

⇒ AB > CD