Question

In the adjoining figure, AB = AD. Prove that BC > CD.

Answer 1

Given:

• A, B, C, D are as in the figure with D on AC (A, D, C are collinear) and AB = AD.

To Prove:

• BC > CD

Proof [Step-wise]:

1. In triangle ABD,

AB = AD,

So, triangle ABD is isosceles and the base angles are equal: ∠ABD = ∠BDA.

2. At vertex B of triangle BCD, the angle ∠CBD equals the difference between ∠ABC and ∠ABD because ∠ABC is made up of ∠ABD and ∠DBC. 

Hence, ∠CBD = ∠ABC – ∠ABD.

3. At vertex D, since A, D, C are collinear, the ray DC is opposite to DA.

Therefore, the interior angle ∠BDC of triangle BCD equals the supplement of ∠BDA: ∠BDC = 180° – ∠BDA. 

Using step 1 (∠BDA = ∠ABD), we get

∠BDC = 180° – ∠ABD

4. Compare ∠CBD and ∠BDC:

∠CBD = ∠ABC – ∠ABD < 180° – ∠ABD

= ∠BDC   ...(Because ∠ABC < 180°)

Thus, ∠BDC > ∠CBD.

5. In triangle BCD, the side opposite the larger angle is the larger side.

Since ∠BDC > ∠CBD, the side opposite ∠BDC (which is BC) is longer than the side opposite ∠CBD (which is CD). 

Hence, BC > CD.