In `DeltaABC, AB = AC` and D is a point in BC so that `BD=CD`. Prove that AD bisects `angleBAC`.

In Figure,AC>AB and D is the point on AC such that AB=AD. Prove that BC>CD

In a DeltaABC , AB = AC and D is a point on AB, such that AD = DC = BC. Then, angle BAC is

In Figure,AD=AE and D and E are points on BC such that BD=EC .Prove that AB=AC

D is a point on the side BC of a DeltaABC such that AD bisects angle BAC . Then

In the adjoining figure, ABC is an isosceles triangle in which AB = AC. Also, D is a point such that BD = CD. Prove that AD bisects /_ A and /_D

In an isosceles DeltaABC with AB = AC, D and E are point on BC such that BE = CD. Show that AD = AE.

If D is a Point on the side BC of a Delta ABC such that AD bisects angleBAC . Then

In /_ABC,AB=AC and D is any point on BC , prove that AB^(2)-AD^(2)=BD.CD