Question

In triangle ABC, AD ⊥ BC. Prove that (AB + AC)(AC – AB) = (CD + BD)(CD – BD).

Answer 1

We are given:

• In triangle △ABC, AD ⊥ BC

• Need to prove: (AB + AC)(AC – AB) = (CD + BD)(CD – BD)

The right-hand side is of the form:

(x + y)(x – y) = x² – y²

So, this identity tells us:

(AB + AC)(AC – AB) = AC² – AB²

(CD + BD)(CD – BD) = CD² – BD²

Apply Pythagoras Theorem in triangles:

In △ABC, with AD ⊥ BC, consider two right-angled triangles:

1. In △ABD: AB² = AD² + BD²

⇒ BD² = AB² – AD²   ...(1)

2. In △ACD: AC² = AD² + CD²

⇒ CD² = AC² – AD²   ...(2)

Subtract (1) from (2):

CD² – BD²

= (AC² – AD²) – (AB² – AD²)

= AC² – AB²

Final Step:

So, (CD + BD)(CD – BD)

= CD² – BD²

= AC² – AB²

(AB + AC)(AC – AB) = AC² – AB²

(AB + AC)(AC – AB) = (CD + BD)(CD – BD)

Hence, Proved.