Question

Abc is a right angled isosceles triangle, right angled at b. Ap the bisector of angle bac intersects bc at p. Prove that AC square = ap square + 2(1 + root 2) bp square.

Answer 1

SInce AP is the angle bisector of angle A.

BP/PC = AB/AB root 2

So PC = BP root 2

Now AB2 = AP2 - BP2

AND AB2 = AC2 - BC2

Equating these two

AP2 - BP2 = AC2 - BC2

AP2 - BP2 = AC2 - 3 BP2 - 2 BP2 root2

By LHS and RHS u will get

AC2 = AP2 + 2(1+root2) BP2