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Question
If P and Q are the points on side CA and CB respectively of ΔABC, right angled at C, prove that (AQ2 + BP2) = (AB2 + PQ2)
Solution
Using the Pythagoras theorem in ΔABC,
ΔACQ, ΔBPC, ΔPCQ, we get
AB2 = AC2 + BC2 ..(1)
AQ2 = AC2 + CQ2 ...(2)
BP2 = PC2 + BC2 ..(3)
PQ2 = PC2 + CO2 ...(4)
Adding the equations (2) and (3) we get
AQ2 + BP2 = AC2 + CQ2 + PC2 + BC2
=(AC2 + BC2) + (CQ2 + PC2)
= AB2 + PQ2
As L.H.S = AQ2 + BP2
= AB2 + PQ2 = R.H.S
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