If A,B,C and D are angles of quadrilateral and `sin(A)/(2)sin(B)/(2)sin(C)/(2)sin(D)/(2)=(1)/(4)`, prove that A=B=C=D=`pi//2`

Explanation

`(2sin(A)/(2)sin(B)/(2))(sin(C)/(2)sin(D)/(2))=1`
`rArr{cos((A-B)/(2))-cos((A+B)/(2))}{cos((C-D)/(2))-cos((C+D)/(2))}=1`
Since `A +B=2pi-(C+D)`, the above equation becomes.
`rArr{cos((A-B)/(2))-cos((A+B)/(2))}{cos((C-D)/(2))+cos((A+B)/(2))}=1`
`rArr((A+B)/(2))-cos((A+B)/(2)){cos((A-B)/(2))-cos((C-D)/(2))}+1-cos((A-B)/(2))cos((C-D)/(2))=0`
This is a quadratic equation is `cos((A+B)/(2))` which has real roots.
`rArr{cos((A-B)/(2))-cos((C-D)/(2))}^(2)-4{1-cos((A-B)/(2))cos((C-D)/(2))}ge0`
`(cos(A-B)/(2)+cos(C-D)/(2))^(2)ge4`
`rArrcos(A-B)/(2)+cos(C-D)/(2)ge2`, now both `cos(A-B)/(2)` and `cos(C-D)/(2)le1` ltbr. `rArrcos(A-B)/(2)=1 &cos(C-D)/(2)=1`
`rArr(A-B)/(2)=0=(C-D)/(2)`
`rArrA=B,C=D`.
Similarly `A=C,B=DrArrA=B=C=D=pi//2`