In figure `angleABD=angleACD, AD` is bisectyor of `angleBAC and AD` meets BC at D. Prove that D is mid-point of BC.

In a tringle ABC, AB = AC and bisector of angle A meets BC at D. Prove that : (i) DeltaABD cong DeltaACD (ii) AD is perpendicular to BC.

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

AB and CD are the parallel sides of a trapeziuml. E is the mid-point of AD. A line through E and parallel to side AB meets the line BC at point F. Prove that F is the mid-point of BC.

In ∆ABC, AD is the bisector of ∠A meeting BC at D, CF ⊥ AB and E is the mid-point of AC. Then median of the triangle is

In the figure, if angleBAC = 90^(@) and AD bot BC , then:

In DeltaABC, angleC is obtuse. The exteral bisector of angleA and angleB meet the extended line BC and AC at point D and E. If AB = AD = BE then find angleACD ?

In the given figure BX is the angle bisector of /_ABC,E is mid point of AD and D is the mid point of BC then (BE)/(EX) is [Diagram not to scale]

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

In the adjoining figure, in DeltaABC , point D is on side BC such that, angleBAC = angleADC . Prove that, CA^2 = CB xx CD .

In triangle ABC, AB gt AC and D is any point on BC. Prove that, AB gt AD.