In triangle ABC, AB=AC and BD is perpendicular to AC. Prove that BC²=2AC.CD

Given: BD is perpendicular to AC and AB = AC

To prove: BC² = 2 AC.CD

Proof:

In Δ BCD by pythagoras theorem, we have

⇒ BC² = BD² + CD² ... (1)

Again, in Δ ABD by pythagoras theorem

AB² = BD² + AD²

⇒ BD² = AB² - AD²

On putting value of BD² in (1), we get

BC² = AB² - AD² + CD² 

⇒ BC² = AB² - (AC - CD)² + CD²

⇒ BC² = AB² - AC² - CD² + 2AC. CD + CD²

⇒ BC² = 2AC. CD [Given, AB = AC. ⇒ AB² = AC²]

[Hence proved]