Question

In ΔABC, ∠B = 90° and D is the mid-point of BC. Prove that AC² = AD² + 3CD².

Answer 1

Given,

In ∆ABD,

By Pythagoras theorem

AD² = AB² + BD²

AB² = AD² – BD²   ...(i)

In △ABC,

AC² = AB² + BC²

AB² = AC² – BC²   ...(ii)

Equate equation (i) and (ii)

⇒ AD² – BD² = AC² – BC²

⇒ AD² – BD² = AC² – (BD + CD)²

⇒ AD² – BD² = AC² – [BD² + CD² + 2(BD)(CD)]

⇒ AD² – BD² = AC² – BD² – CD² – 2(BD)(CD)

Since D is the midpoint, then, BD = CD.

⇒ AD² – BD² = AC² – BD² – CD² – 2(CD)(CD)

⇒ AD² = AC² – `\cancel("BD"^2)` + `\cancel("BD"^2)` – CD² – 2CD²

⇒ AD² = AC² – 3CD²

⇒ AC² = AD² + 3CD²

Hence, proved.