Question

P is any point inside ΔABC and PQ, PR and PS are drawn perpendiculars to sides AB, BC and AC respectively. Prove that AQ² + BR² + CS² = BQ² + RC² + AS².

Answer 1

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.

Applying Pythagoras theorem on all the right angle triangles, we get

In △AQP, AQ² = PA² – PQ²   ...(i)

In △BRP, BR² = BP² – PR²   ...(ii)

In △CSP, CS² = CP² – PS²   ...(iii)

Adding equations (i), (ii) and (iii), we get

AQ² + BR² + CS² = PA² + PB² + PC² – PQ² – PR² – PS²

AQ² + BR² + CS² = (PA² – PS)² + (PB² – PR²) + (PC² – PQ²)

AQ² + BR² + CS² = AS² + BQ² + RC²

Hence, proved.