Two straight lines are drawn through any point X and exterior to a circle to intersect the circle at points A, B and points C, D respectively . Prove that `DeltaXACandDeltaXBD` are equiangular.

Explanation

ABCD is a cyclic quadrilateral.
`thereforeangleABD+angleACD=180^(@)` …………….(1)
Again , `angleXCA+angleACD=1"straight angle" =180^(@)` ………..(2)
Then , subtracting (1) from (2) we get `angleXCA-angleABD=0" or " , angleXCA=angleABD` .
Again in cyclic quadrilateral ABDC , `angleBAC+angleBDC=180^(@)` .........(3)
and `angleXAC+angleBAC=1"straight angle"=180^(@)` ........(4) ltbr? Then subtracting (3) and(4) we get,
`angleXAC-angleBDC=0 " or " , angleXAC=angleBDC`
`therefore` in triangles `DeltaXACandDeltaXBD`,
`angleXCA=angleXBD[becauseangleABD=angleXBD]`
and `angleXAC=angleXDB`
two angles of each of `DeltaXACandDeltaXBD` are equal.
Hence `DeltaXACandDeltaXBD` are equianhular.