AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

AD bisects `angleA`

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that AD bisects BC

AD is an altitude of an isosceles DeltaABC in which AB = AC. Show that (i) AD bisects BC, (ii) AD bisects angleA .

In the adjoining figure, DeltaABC is an isosceles triangle in which AB = AC and AD is the bisector of angleA . Prove that: DeltaADB~=DeltaADC

In the adjoining figure, DeltaABC is an isosceles triangle in which AB = AC and AD is the bisector of angleA . Prove that: BD = CD

In the adjoining figure, DeltaABC is an isosceles triangle in which AB = AC and AD is the bisector of angleA . Prove that: ADbotBC

In the adjoining figure, DeltaABC is an isosceles triangle in which AB = AC and AD is the bisector of angleA . Prove that: angleB=angleC

AD and BE are respectively altitudes of an isosceles triangle ABC with AC=BC . Prove that AE=BD

In the figure, ABC is an isosceles triangle in which AB = AC and LM is parallel to BC. If angleA= 50^(@) find angleLMC .

In the adjoinng figure,Delta ABC is an isoceless triangle in which AB=AC .Also,D is a point such that BD=CD. Prove that AD bisects /_A and /_D.

In the adjoining figure, DeltaABC is an isosceles triangle in which AB = AC and AD is a median. Prove that: angleBAD=angleCAD