`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

Since points `B,C,D and E` are Concyclic. Sum of opposite angles of cyclic quadrilateral is `180^@`.
So, `∠1 + ∠3=180°`
and, `∠22 + ∠24=180°`
Also, AABC is an Isosceles Triangle.
AB=AC(Given)
`∠3=∠4`→when opposite sides are equal angle opposite to them are equal.
Gives, `∠1+∠4=180^@`
`∠2+∠3=180^@`
But these are interior angles on the same side of transversal B D and EC. So, if sum of interior angles on the same side of transversal is `180^@`, then lines are parallel.
DE || BC
`∠4=∠5'
'∠3=∠6`
when lines are parallel, corresponding angles are equal.
But, `∠3=∠4`
Which gives, `∠5=∠6`
AD=A E→If opposite angles in a triangle are equal, side opposite to them are equal.
Draw, AM⊥DE
In ΔAMD and ΔΑΜΕ
AD=AE→Proved above
∠AMD=∠AME→each being `90^@`.
AM is common.
△AMD = △AME→[R HS congruency rule]
→DM ME→[CPCT]
Join ,AO,which will pass 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 because, DM = ME.