If in triangle ABC, AD is a median and AM is perpendicular to BC then prove that AB²= AD² – BC × DM + 1/4 BC².

Given: D is the mid-point of BC.

And AM ⊥ BC.

To Prove:

Proof: In right-angled ∆ABM,

By Pythagoras theorem, we can write as

AB² = AM² + BM²

[∵ (hypotenuse)² = (perpendicular)² + (base)²] …(i)

In right-angled ∆AMD,

By Pythagoras theorem, we can write as

AD² = AM² + MD²

[∵ (hypotenuse)² = (perpendicular)² + (base)²] …(ii)

Subtract equation (ii) from equation (i), we get

AB² – AD² = AM² – AM² + BM² – MD²

⇒ AB² – AD² = 0 + BM² – MD²

⇒ AB² = AD² – MD² + BM²

⇒ AB² = AD² – DM² + (BD – DM)²

⇒ AB² = AD² – DM² + BD² + DM² – 2(BD)(DM)

⇒ AB² = AD² + BD² – 2(BD)(DM)

[∵ ]

Hence, proved.