Question

Let ABC be a triangle with AB > AC. Let P be a point on the line AB beyond A such that AP + PC= AB. Let M be the midpoint of BC and let Q be the point on the side AB such that CQ ⊥ AM .Prove that BQ = 2AP.

Answer 1

Extend BP to F such PF =PC. Then AF = AP + PF = AP + PC= AB. Hence A is the mid-point of BF.Since M is the mid-point of BC, it follows that AM || FC. But AM ⊥ CQ. Hence FC ⊥ CQ at C. Therefore QCF is a rightangled triangle. Since PC=PF, it follows that ∠PCF = ∠PFC. Hence ∠PQC = ∠PCQ which gives PQ = PC = PF. This implies that P is the mid - point of QF.

Thus we have AP + AQ = PE and BQ + QA = AP + PF. This gives

2AP + AQ = PE + AP = BQ + QA

We conclude that BQ = 2AP.