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 + 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