`A B C D` is a parallelogram. The circle through `A ,Ba n dC` intersects `C D` produced at `E ,` prove that `A E=A Ddot`

Explanation

We conclude from the figure,
`=>/_ADE+/_ADC=180^@` (Due to Linear pair) ....... `(i)`
Also, `squareABCE` is a cyclic quadrilateral.
So, `/_AED+/_ABC=180^@` ....... `(ii)` (opposite angles of cyclic quadrilateral sum upto `180 ^@`)
`/_ADC=/_ABC` ....... `(iii)` (Opposite angles of a parallellogram are equal)
Using `(i), (ii)` and `(iii)`,

`/_AED+/_ABC=/_ADE+/_ABC`
`=>/_AED=/_ADE`

Now, In
`/_\AED,`
`/_AED=/_ADE`
`=>AD=DE` (Sides opposite to equal angles)
Hence proved.