In the given figure, `Delta ABC` is isosceles with `AB= AC` and `/_ ABC = 50^(@)` . Find `/_ BDC ` and `/_ BEC`.

Explanation

`AB = AC implies /_ACB = /_ ABC = 50^(@)`
But, `/_ ABC + /_ ACB + /_BAC = 180^(@)`
`implies 50^(@) +50^(@) + /_ BAC = 180^(@)`
`implies /_ BAC =180^(@) -100^(@) = 80^(@) `
` :. /_ BDC = /_ BAC = 80^(@)` `[ /_ s` in the same segment ].
Now, BECD being a cyclic quadrilateral, we have
`/_ BEC + /_ BDC = 180^(@)`
`implies /_ BEC + 80^(@) = 180^(@)`
`implies /_ BEC = 180^(@)-80^(@) = 100^(@)`
Hence, `/_ BDC = 80^(@)` and `/_ BEC = 100^(@)`.