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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.
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Solution
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
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