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 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.

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, AB > AC. E is the mid-point of BC and AD is perpendicular to BC. Prove that : AB^2 +AC^2=2AE^2 +2BE^2

In the figure, BX bisects angle ABC and intersects AC at point D. Line segment CY is perpendicular to AB and intersects BX at point P. If Y is mid-point of AB, prove that : point P is equidistant from A and B.

In triangle ABC, AB=AC and BD is perpendicular to AC. Prove that : BD^(2)-CD^(2)=2CDxxAD

In a triangle ABC, D is a point on BC such that AD is the internal bisector of angle A . Let angle B = 2 angle C and CD = AB. Then angle A is

ABC is a right triangle with AB = AC .If bisector of angle A meet BC at D then prove that BC = 2 AD .

In a triangle ABC, AD is a medium and E is mid-point of median AD. A line through B and E meets AC at point F.

In a triangle ABC,D is the mid - point of BC, AD is produced upto E so that DE = AD. Prove that : AB=EC