Bisector of angle B of triangle ABC intersects side AC at point P, then point P is :

1. equidistant from vertices A and C

2. PA = PB

3. PB = PC

4. equidistant from sides AB and BC

Given: AX bisects angle BAC and PQ is perpendicular bisector of AC which meets AX at point Y.

Prove :

(i) X is equidistant from AB and AC.

(ii) Y is equidistant from A and C.

Given: PQ is a perpendicular bisector of side AB of the triangle ABC.

Prove: Q is equidistant from A and B.

Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.