The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F. Prove that : F is equidistant from AB and AC.

The perpendicular bisectors of the sides of a triangle ABC meet at I. Prove that : IA = IB = IC.

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.

In triangle ABC, D is mid-point of AB and CD is perpendiicular to AB. Bisector of angleABC meets CD at E and AC at F. Prove that: E is equidistant from A and B

In triangle ABC, D is mid-point of AB and CD is perpendiicular to AB. Bisector of angleABC meets CD at E and AC at F. Prove that: F is equidistant from AB and BC.

The equation of perpendicular bisectors of side AB,BC of triangle ABC are x-y=5 , x+2y=0 respectively and A(1,-2) then coordinate of C

Show that the perpendicular bisectors of the sides of a triangle are concurrent.

In each of the given figures, PA = PB and QA = QB. Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.

Perpendicular bisectors of the sides AB and AC of a triangle ABC meet at O. What do you call the point ?

Construct a triangle ABC which angle ABC=75^(@), AB=5" cm"and BC =6.4" cm" . Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisector intersect each other at point P, prove that P is equidistant from B and C, and also from AC and BC.