Question

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

Answer 1

Given: A point P lies in the angle ABC and $PL\perp BAandPM\perp BCandPL=PM.PB$ is joined

To prove : PB is the bisector $\angle ABC,$

Proof : In right $\Delta PLBand\Delta PMB$

Side PL = PM (Given)

Hyp. PB = PB (Common)

$\therefore \Delta PLB\cong \Delta PMB\text{(RHS axiom)}$

$\therefore \Delta PBL=\Delta PBM(c.p.c.t.)$

$\therefore PB\text{is the bisector of}\angle ABC$