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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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प्रश्न

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. 

योग
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उत्तर

  
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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अध्याय 16: Loci (Locus and Its Constructions) - Exercise 16 (A) [पृष्ठ २३७]

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सेलिना Concise Mathematics [English] Class 10 ICSE
अध्याय 16 Loci (Locus and Its Constructions)
Exercise 16 (A) | Q 8. | पृष्ठ २३७

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