In `DeltaABC`, the internal bisects BD, CD of angle ABC, angle ACB respectively meet at D.show that angle `BDC = 90^@+A/2`

In Delta ABC, the internal bisects BD,CD of angle ABC,angle ACB respectively meet at D.show that angle BDC=90^(@)+(A)/(2)

angleBAC : angle ABC : angle ACB = what ?

In triangle ABC, AD is perpendicular to side BC and AD^2 = BD xx DC . Show that angle BAC = 90^@ .

The bisector of the interior angles angleB and angle C of the DeltaABC , meet at H. Prove that angleBHC=90^@+(1)/(2)angleBAC .

In Fig. 10.38, angle ABC = 69^@ , angle ACB = 31^@ , find angle BDC .

If in a triangle ABC,CD is the angular bisector of the angle ACB then CD is equal to