Question

In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.

Answer 1

Steps of Construction:

i) In the given triangle, draw the angle bisector of ∠BAC.

ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.

P is the required point which is equidistant from AB and AC as well as from B and C.

Since P lies on angle bisector of ∠BAC,

It is equidistant from AB and AC.

Again, P lies on perpendicular bisector of BC,

Therefore, it is equidistant from B and C.