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

**(i) **

**(ii) **BP=BQ or B is equidistant from the arms of .

**Solution (i)**

In and , we have

**(It is given that AB is the bisector) (1)**

** (Common) (2)**

** (Each equal to ) (3) **

Therefore, **by AAS congruence rule**, we have .

**Solution (ii)**

From **solution (i), **we know that .

Therefore, BP=BQ ** (Corresponding parts of congruent triangles are equal)**