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In an isosceles triangle A B C with A...

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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AB`=` AC
`implies /_ ABC = /_ ACB ` ….(i)
Side BD of the cyclic quadrilateral BCED is produced to A.
`:. /_ ADE = /_ ACB ` …(ii)
[ext. `/_ ADE =`int. opp. `/_ C ]` .
From (i) and (ii) , we get
`/_ ABC = /_ ADE.`
But, these are corresponding angles.
Hence,`DE || BC`
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