Question

In an isosceles triangle ABC with AB=AC BD is perpendicular from B to the side AC. Prove that BD2-CD2= 2CD.AD

Answer 1

Solution:

We have to prove BD² -CD² = 2CD×AD

Proof:

In Right angled triangle ABD, now by using Pythagoras theorem we have  AB² = AD²+BD²

and AB = AC

⇒AC² = AD²+BD²

⇒(AD+DC)² = AD²+BD²

⇒AD²+DC² +2AD×DC = AD²+BD² , ∵ (a+b)² = a²+b²+2ab

⇒2AD×DC = BD²-DC²

Hence proved

BD²-DC² = 2AD×CD