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प्रश्न
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)
उत्तर
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 + CQ2 .......(4)
Adding the equations (2) and (3) we get
AQ2 + BP2 = AC2 + CQ2 + PC2 + BC2
= (AC2 + BC2) + (CQ + PC2)
= AB2 + PQ2
As
L.H.S. = AQ2 + BP2
= AB2 + PQ2
= R.H.S
Hence Proved
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