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

Explanation

It is given that ABCE is a cyclic quadrilateral.
`:. /_ ABC + /_ AED = 180^(@)` .....(i)
Also, EDC is a straight line.
`:. /_ ADE + /_ ADC = 180^(@)`
But,`/_ ADC = /_ABC ` [ opposite `/_ s ` of a `||` gm ]
`:. /_ ADE + /_ABC =180^(@)` ...(ii)
Thus,`/_ ABC + /_ AED = /_ ADE + /_ ABC` [ from (i) and (ii) ]
`implies //_AED = /_ADE`
`implies AD =AE ` [ opposite sides of equal of `/_ s` of `Delta ` ADE ]
Hence, `AD = AE`.