In a `!ABC` bisector of angle C meets the side AB at D and circumcricle at E. The maximum value of CD . DE is equal to

In Delta ABC with fixed length of BC, the internal bisector of angle C meets the side AB at D and meet the circumcircle at E. The maximum value of CD xx DE is

In A B C with fixed length of B C , the internal bisector of angle C meets the side A Ba tD and the circumcircle at E . The maximum value of C DxD E is c^2 (b) (c^2)/2 (c) (c^2)/4 (d) none of these

In a triangle ABC, angle bisector of angle BAC cut the side BC at D and meet the circumcircle of Delta ABC at E , then find AB .AC+DE. AE.

In a triangle ABC, angle bisector of angleBAC cut the side BC at D and meet the circumcircle of DeltaABC at E. If AC = 4 cm, AD = 5 cm, DE = 3 cm. Find the length of AB.

In triangle ABC,angleC=(2pi)/3 and CD is the internal angle bisector of angleC meeting the side AB at D. If Length CD is 1, then H.M. of a and b is equal to : (a) 1 (b) 2 (c) 3 (d) 4

In a triangle ABC, angle bisector of angleBAC cut the side BD and meet the circumcircle of DeltaABC at E. If AC = 4 cm, AD = 5 cm, DE = 3 cm. Find the length of AB.