I have drawn a right angled triangle ABC whose ∠A is right angle. I took two points P and Q on the sides AB and AC respectively. By joining P,B,Q and C, P let us prove that, BQ² + PC² = BC² + PQ²

Given: ABC is a right angled triangle with ∠A = 90°

To prove: BQ² + PC² = BC² + PQ²

Proof:

By applying Pythagoras theorem in ΔAPQ, we get,

PQ² = AP² + AQ² ……………… (1)

By applying Pythagoras theorem in ΔABQ, we get,

BQ² = AB² + AQ² ……………… (2)

By applying Pythagoras theorem in ΔAPC, we get,

PC² = AP² + AC² ………………… (3)

By applying Pythagoras theorem in ΔABC, we get,

BC² = AB² + AC² ……………… (4)

By adding (1) and (2) we get,

BQ² + PC² = AB² + AQ² + AP² + AC²

⇒ BQ² + PC² = AB² + AC² + AQ² + AP²

Substituting from (1) and (4) we get,

BQ² + PC² = BC² + PQ²

Hence proved.