`D` and `E` are points on equals sides `AB` and `AC` of an isosceles `Delta ABC` such that `AD` `=` `AE` . Prove that the points of `B,C,E` and `D` are concyclic.

Explanation

In order to prove that the points `B,C,E` and `D` are concyclic, we must show that `/_ ECB + /_ EDB = 180^(@)`
Now, `AB = AC implies /_ ABC = /_ ACB` …..(i)
And, ` AD = AE implies /_ AED = /_ ADE ` ….(ii)
We know that the sum of the angles of a triangles is `180^(@)`.
`:. ` in `Delta ABC` and `Delta ADE`, we have
`/_ A + /_ ABC + /_ ACB = 180^(@)` and `/_ A+ /_ADE + /_AED = 180^(@)`
`implies /_A + 2/_ACB = 180^(@)` and `/_ A + 2/_ ADE = 180^(@)` [using (i) and (ii)]
`implies /_A+2/_ ACB= /_ A+2/_ADE`
`implies 2/_ ACB = 2/_ ADE implies /_ ACB = /_ ADE ` .....(iii) `[: ' /_ ACB = /_ECB]`
`:. /_ECB+/_EDB= /_ADB+/_EDB` [ using (iii) ]
`= 180^(@) [ :' /_ ADE +/_EDB=` a staright `/_ ]`
Thus, `/_ECB + /_EDB = 180^(@) `,
This shows that the points `B,C,E` and `D` are concyclic.