Question

In ΔABC, ∠ABC is an acute angle. Prove that : AC² = AB² + BC² – 2BC.BD.

Answer 1

Given: In ΔABC, ∠ABC is acute and AD ⟂ BC.

D is the foot of the perpendicular from A to BC.

To Prove: AC² = AB² + BC² – 2 × BC × BD

Proof [Step-wise]:

1. In right triangle ABD right at D.

By Pythagoras:

AB² = AD² + BD²

2. In right triangle ACD right at D.

By Pythagoras:

AC² = AD² + DC²

3. Note that DC = BC − BD.

4. Substitute DC in step 2:

AC² = AD² + (BC – BD)²

= AD² + BC² – 2 × BC × BD + BD²

5. Replace AD² + BD² from step 1 by AB² to get:

AC² = AB² + BC² – 2 × BC × BD

Hence, AC² = AB² + BC² – 2 × BC × BD, as required.