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`.

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 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 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 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, ACgtAB and AD is the bisector of angleA. show that angleADC gt angleADB .

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.

AB is a line segment and line 1 is its perpendicular bisector.If a point P lies on 1 show that P is equidistant from A and B .

In the adjacent figure, the bisectors of angleA and angleB meet in a point P. If angleC=100^(@) and angleD=60^(@) , find the measure of angle APB .