Question

P is any point in the interior of angle ABC such that it is equidistant from AB and BC. Prove that P lies on the bisector of angle ABC

Answer 1

Step-by-step explanation:

Given:P is in the interior of <ABC

AP=PB

TO PROVE :P lies on the bisector of<ABC

=<ABP=<PBC

Construction: Take point X on ray A and Y on B,join point X to P and Y to P

Solution: Now 2 ∆ are formed

In ∆XBP and ∆YBP

BP=BP...........................(1) Common side

XP = YP.........................(2) given

<BXP = <BYP.................(3)

∆XBP=∆YBP ...(from 1,2 and 3)

<XBP=<YBP (CACT) proved

Answer 2

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