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 .

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

Line l is the bisector of an angle angleA and B is any point on li. BP and BQ are perpendicular forms B to the arms of angleA . Show that: triangleAPB ~= triangleAQB (ii) BP = BQ or B is equidisitant 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 angleA and B is any point on l. BP and BQ are perpendiculars from B to the arms of angleA show that: triangleAPBequivtriangleAQB

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 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 angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of angle A (see Fig. 7.20). Show that: (i) triangle APB cong triangle AQB (ii) BP = BQ or B is equidistant from the arms of angle A .

Ray 1 is the bisector of an angle angle A and B is any point on I. BP and BQ are perpendiculars from B to the arms of angle A (see the given figure). Show that: (i) triangleAPB=triangleAQB (ii) BP = BQ or B is equidistant from the arms of angleA