`D\ a n d\ E` are, respectively, the points on equal sides `A B\ a n d\ A C` of an isosceles triangle `A B C` such that `B ,\ C ,\ E ,\ a n d\ D` are concyclic as shown in Figure. If `O` is the point of intersection of `C D\ a n d\ B E ,` prove that `A O` is the bisector of line segment `D E`

Explanation

Answer
Points `B,C,D and E` are Concyclic.
Then, Sum of opposite angles of cyclic quadrilateral is `180^@`
So,`/_1+/_3=180^@`
and `/_2+/_4=180 ^@`
Also, `/_\ABC` is the Isosceles triangle.
`AB=AC`(Given)
`/_3=/_4` [opposite sides are equal angle opposite to them are equal.]
Given:`/_1+/_4=180^@`
and
`/_2+/_3=180^@`
Sum of interior angles on the same side of transversal is `180^@` , then lines are parallel.
` ..' DE|\ |BC`
`/_4=/_5`
and
`/_3=/_6`
when lines are parallel, corresponding angles are equal.
But,`/_3=/_4`

`=>/_5=/_6`
`:.AD=AE` [opposite angles in a triangle are equal, the side opposite to them are equal.]
From the figure
`AM_|_DE`
In `/_\AMD` and `/_\AME`
`AD=AE` proved above.
`/_AMD=/_AME` each being `90^@`
`AM` is common.
`/_\AMD~=/_\AME` by `RHS` congruency rule
`=>DM=ME` by `C.P.C.T`
From the fig.`AO` which passes through `M`, as `/_AME+/_OME=180^@` ,showing points `A,M` and `O` are collinear.
which shows, Segment `AO` is the bisector of line segment `DE`
`..'DM=ME`.
Hence proved.