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

Explanation

`/_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 `