Line l is the bisector of an `angleA` and B is any point on I. BP and BQ are perpendiculars from B to the arms of `angleA` (see figure). Show that:

BP=BQ or B is equidistant from the arms of `angleA`.

Line l is the bisector of an angleA and B is any point on I. BP and BQ are perpendiculars from B to the arms of angleA (see figure). Show that: DeltaAPB~=DeltaAQB

In the given figure, line I is the bisector of an angle A and B is any point on I. BP and BQ are perpendiculars from B to the arms of angleA . Show that B is equidistant from the arms of angleA .

In Figure, line l is the bisector of angle A and B is any point on l . B P and B Q are perpendiculars from B to the arms of Adot Show that : A P B~=A Q B B P=B Q or B is equidistant from the arms of /_A . Figure

In Figure, line l is the bisector of angle A a n d B is any point on ldotB P a n d B Q are perpendiculars from B to the arms of Adot Show that: A P B ~= A Q B BP=BQ or B is equidistant from the arms of /_A

line l is the bisector of an angle /_A\ a n d/_B is any point on l. BP and BQ are perpendiculars from B to the arms of /_A . Show that: (i) DeltaA P B~=DeltaA Q B (ii) BP = BQ or B is equidistant from the arms of /_A

In the given figure, line l is the bisector of an angle angleA and B is any point on l. If BP and BQ are perpendiculars from B to the arms of angleA , show that (i) DeltaAPB~=DeltaAQB (ii) BP = BQ, i.e., B is equidistant from the arms of angleA .

In DeltaABC , the bisector AD of angleA is perpendicular to side BC (see figure). Show that AB =AC or DeltaABC is isosceles.

P is any point in the angle ABC such that the perpendiculars drawn from P on AB and BC are equal. Prove that BP bisects angle ABC.

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

AB is a line segment and line l is its perpendicular bisector.If a point P lies on l show that P is equidistant from A and B.