`A B C D` is a cyclic quadrilateral. `A Ba n dD C` are produced to meet in `Edot` Prove that `E B C-E D Adot`

Explanation

ABCD is a cyclic quadrilateral and AB and CD are produced to meet at E.
We know that opposite angles of a cyclic quadrilateral are supplementary.
∴ `∠BAD+∠BCD=180^@` ------ ( 1)
and `∠BAD+∠EAD=180@`----- ( 2 )
From equation ( 1 ) and ( 2 ), we get,
`∠BAD+∠EAD=∠BAD+∠BCD`
∴ ` ∠EAD=∠BCD` --- ( 3 )
Similarly,
⇒ `∠ABC+∠ADC=180^@` ---- ( 4
and `∠ADC+∠ADE=180^@`----- ( 5 )
From equation ( 4 ) and ( 5 ), we get
⇒ `∠ADC+∠ADE=∠ADC+∠ABC`
⇒ `∠ADE=∠ABC `--- ( 6 )
And `∠AED=∠BEC` [ Common angle ] ---- ( 7 )
∴ From equation ( 3 ), ( 6 ) and ( 7 ),`△EBC` and `△EDA` are equiangular.