Question

In the given figure, triangle ABC is a right-angled at B. D is the mid-point of side BC. Prove that AC² = 4AD² – 3AB².

Answer 1

Given: ∠ABC = 90°, D is mid-point of side BC.

To prove: AC² = 4AD² – 3AB².

Proof: Since D is mid-point of side BC,

∴ BD = DC = `1/2` BC  ......(i)

Now, in ΔABC, by Pythagoras theorem,

AB² + BC² = AC²

⇒ AB² + (2BD)² = AC²  .....[Using (i)]

⇒ AB² + 4BD² = AC²   ......(iii)

Similarly, in ΔABD, by Pythagoras theorem,

AB² + BD² = AD²

⇒ BD²  = AD² – AB²

Putting the value of BD² in equation (ii),

AB² + 4(AD² – AB²) = AC²

⇒ AB² + 4AD² – 4AB² = AC²

⇒ 4AD² – 3AB² = AC²

or AC² = 4AD² – 3AB²

Hence proved.