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In an isosceles triangle `A B C` with `A B=A C ,` a circle passing through `B\ a n d\ C` intersects the sides `A B\ a n d\ A C` at `D\ a n d\ E` respectively. Prove that `D E B C`

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Answer
In `/_\ ABC`
`/_B=/_C.....(1)`
In the cyclic quadrilateral `CBDE`, side `BD` is produced to `A`.
We know that exterior angle is equal to opposite interior angle.
i.e `/_D=/_C....(2)`
From `(1)` and `(2)` `/_ADE=/_ABC`
So, corresponding angles are equal
Hence, `DE|\ |BC`
Hence proved.
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