In perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.

Explanation

Let P be a point within `angle ABC` such that PM = PN. We have to provethatP lies on the bisector of `angle ABC` i.e.,` angle 1 = angle 2.`
`In triangles PMB and PNB`
`PM = PN ` ( given )
`BP = BP` ( common )
`angle PNB = angle PMB` ( right angle )
Therefore,` triangle PMB = triangle PNB`
`angle 1 = angle 2` [ `therefore` angle B = angle C ]
Hence, P lies on the bisector of angle ABC.