ABCD is a cyclic quadrilateral .The chord DE is the external bisector of `angleBDC` Prove that AE (or extended AE), is the external bisector of `angleBAC`.

Explanation

Two angles in circle produced by the arc BC are `angleBAC and angleBDC` .
`thereforeangleBAC=angleBDC`………(1)
Again , two angles in circle produced by the arc CE arc `angleCAEand angleCDE`.
`thereforeangleCAE=angleCDE` ………….(2)
Now , `angle FDE+angleCDE+angleBDC=1" straight angle "=180^(@)`
or, `angleCDE+angleCDE+angleBDC=180^(@)[becauseDE" is bisector of "angleFDCthereforeangleFDE=angleCDE]`
or, `2angleCDE+angleBDC=180^(@)`.......(3)
Again , `angle GAE+angleCAE+angleBAC=1 " straight angle"=180^(@)` ........(4)
From (3) and (4) we get , 2 `angleCDE+angleBDC=angleGAE+angleCAE+angleBAC`
or, `2angleCDE=angleGAE+angleCAE` [from (1)]
or, `2angleCAE=angleGAC` [by the figure]
or, `angleCAE=(1)/(2)angleGAC`
`therefore` AE is the bisector of `angleGAC` Hence AE is the external bisector of `angleBAC`