Given `:` `square ` ABCD is cyclic. `/_DCE ` is an exterior angle of `square ` ABCD. To Prove `: /_DCE = /_BAD ` Complete the proof by filling the boxes.

Given `:` `square ` ABCD is cyclic. `/_DCE ` is an exterior angle of `

1.	Given `:` `square ` ABCD is cyclic. `/_DCE ` is an exterior angle of `square ` ABCD. To Prove `: /_DCE = /_BAD ` Complete the proof by filling the boxes.
Answer» `/_DCE + /_BCD = square` …(Linear pair of angles ) ….(1) `square ` ABCD is cyclic `/_BAD + square = 180^(@)` …(Theorem of `square ` ) ….2) `:.` from (1) and (2), we get, `/_DCE + square = square + /_BCD ` `:. /_DCE =square ` Activity `:` ` /_ DCE + /_BCD = 180^(@)` ....(Linear pair of angles ) ...(1) `square ` ABCD is cyclic `/_ BAD + /_BCD = 180^(@)` ...(Theorem of Cyclic quadrilateral ) ...(2) `:.` from (1) and (2), we get `/_ DCE + /_BCD = /_BAD + /_BCD ` `:. /_DCE = /_BAD `

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Given `:` `square ` ABCD is cyclic. `/_DCE ` is an exterior angle of `square ` ABCD. To Prove `: /_DCE = /_BAD ` Complete the proof by filling the boxes.

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