If the median AM , angle bisector AD and altitude AH drawn from vertex A of a triangle ABC divide angle A into four equal (D lying between H and M) , then

Options

A. `A=pi/3`
B. `A=pi/2`
C. `(AC)/(AB)=sqrt2+1`
D. `(AC)/(AB)=1/(sqrt2+1)`

Explanation

In `DeltaAMB`, we have
`(AM)/(sin(90^(@)-theta))=(BM)/(sin3theta)rArr(AM)/(BM)=(costheta)/(sin3theta)`

In `DeltaAMC`, we have
`(AM)/(sin(90^(@)-3theta))=(MC)/(sintheta)rArr(AM)/(MC)=(cos3theta)/(sintheta)`
Since AM is the median of `DeltaABC`
`thereforeBM=MCrArr(AM)/(BM)=(AM)/(MC)`
`rArr(costheta)/(sin3theta)/(sintheta)rArrsin6theta-sin2theta`
`rArr6theta=pi-2thetarArrtheta=pi/8rArrangleA=4theta=pi/2`