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

Options

A. AE is HM of b and c
B. `AD=(2bc)/(b+c)cosA/2`
C. `EF=(4b)/(b+c)sinA/2`
D. All of these

Explanation

We have,
Area `DeltaABC` = Area `Delta ABD` + Area `DeltaACD`
`rArr1/2bcsinA=1/2cxxADxxsinA/2+1/2bxxADxxsinA/2`
`rArrAD=(2bc)/(b+c)cosA/2`
So, option (a) is correct.

In `DeltaAED,` we have
`cosA/2=(AD)/(AE)`
`rArrAE=ADsecA/2`
`rArrAE=(2bc)/(b+c)rArr` AE is the HM of b and c.
So, option (b) is correct.
In `DeltaAEF`, we have
`ADbotEf`and AD is the bisector of `angleA`
`thereforeED=DF`
`rArrEF=2DE=2ADtanA/2=2xx(2bc)/(b+c)cosA/2xxtanA/2`
`rArrEF=(4bc)/(b+c)sinA/2`
So, option (c) is correct.
Hence, options (a), (b) and (c) are correct.