Answer :

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

To prove: BQ^{2} + PC^{2} = BC^{2} + PQ^{2}

Proof:

By applying Pythagoras theorem in ΔAPQ, we get,

PQ^{2} = AP^{2} + AQ^{2} ……………… (1)

By applying Pythagoras theorem in ΔABQ, we get,

BQ^{2} = AB^{2} + AQ^{2} ……………… (2)

By applying Pythagoras theorem in ΔAPC, we get,

PC^{2} = AP^{2} + AC^{2} ………………… (3)

By applying Pythagoras theorem in ΔABC, we get,

BC^{2} = AB^{2} + AC^{2} ……………… (4)

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

BQ^{2} + PC^{2} = AB^{2} + AQ^{2} + AP^{2} + AC^{2}

⇒ BQ^{2} + PC^{2} = AB^{2} + AC^{2} + AQ^{2} + AP^{2}

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

BQ^{2} + PC^{2} = BC^{2} + PQ^{2}

Hence proved.

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