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)

Answer 1

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