Answer :
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.
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