In a `DeltaABC, AD` is the bisector of the angle A meeting BC at D. If I is the in centre of the triangle, then `AI: DI` is equal to

In a Delta ABC,AD is the bisector of the angle A meeting BC at D.If I is the in centre of the triangle,then AI:DI is equal to

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 a DeltaABC, AD is angle bisector , AB=7 cm , AC=8 cm, BC=6cm , If I is the incentre . Find AI/ID.

The bisector of angleA of DeltaABC cuts BC at D and the circumcircle of the triangle at E. then

In DeltaABC , angle bisector of angleA intersect BC at D and intersects circumcircle of triangle at E then.

In a Delta ABC, CD is the angle bisector of interior angle C . Which meets AB at D . M is a point on CD such that AD=AM. IF angle B=59^(@) find angle MAC .