Question

In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC. 

Answer 1

  
Construction: From P, draw PL ⊥ AB and PM ⊥ BC

Proof: In ΔPLB and ΔPMB

∠PLB = ∠PMB  ...(Each = 90°)

∠PBL = ∠PBM  ...(Given)

PB = PB   ...(Common)

∴ By Angle – angle side criterion of congruence, 

ΔPLB ≅ ΔPMB  ...(AAS postulate)

The corresponding parts of the congruent triangles are congruent

∴ PL = PM   ...(C.P.C.T.)

Hence, P is equidistant from AB and BC