In ` A B C ,` internal angle bisector of `/_A` meets side `B C` in `D .` `DE_|_A D` meets `A C` at `E and A B` at `F.` Then (a) `A E` is in `H.P.` of `b` and `c` (b) `A D=(2b c)/(b+c)cosA/2` (c) `E F=(4b c)/(b+c)sinA/2` (d) ` AEF` is isosceles

Options

A. AE in H.M of b and c
B. `AD = (2bc)/(b + c) cos.(A)/(2)`
C. `EF = (4bc)/(b +c) sin.(A)/(2)`
D. `DeltaAEF` is isosceles

Answer

A, B, C, D

Explanation

In `DeltaAFE, AD` is angle bisector of `angleA and AD bot EF`.
`:.` D is midpoint of EF and `DeltaAEF` is isosceles triangle.
Therefore, `DeltaAFE` is an isosceles triangle. Now,
To find AD.

`rArr (1)/(2) bc sin A = (1)/(2) c AD "sin" (A)/(2) + (1)/(2) b AD "sin "(A)/(2)`
`rArr AD = (2bc "cos"(A)/(2))/(b+c)`
Also, `AD = AE "cos"(A)/(2) " " ("from " DeltaADE)`
`rArr AE = (2bc)/(b+c) = H.M` of b and c
Again `EF = 2DE = 2 AD "tan"(A)/(2) = (4bc "sin"(A)/(2))/(b+c)`