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Internal bisector of /A of triangle ABC ...

Internal bisector of `/_A` of triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F. If a, b, c represent sides of `DeltaABC`, then

A

AE is H.M of b and c

B

`AD=(2bc)/(b+c)"cos"(A)/(2)`

C

`EF=(4bc)/(b+c)"sin"(A)/(2)`

D

triangle AEF is isosceles

Text Solution

Verified by Experts

The correct Answer is:
(a,b,c,d)
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