Draw a circle with the point a of quadrilateral ABCD as centre which passes through the points B,C and D.Prove that `angle CBD+ angleCDB=(1)/(2) angle BAD`.

Explanation

Let a circle be drawn with centre at A of the quadrilateral ABCD. Which passes through B,C and D.
To prove : We have to prove that `angleCBD+ angle CDB=(1)/(2) angle BAD`
Construction : Let us join `A, B, B, C,C, D, D, A,B, D and C,A`.
Proof : he central angle produced by the are `overset(frown)(BC)= angle BAC` and angle in circle `angleBDC`
`:. BDC=(1)/(2) angle ABCD.....(1)` [ by ptheorem-34]
Again, the central angle produced by `overset(frown)(CD)= angle CAD` and angle in cricle `=angleCBD`
`:. angle CBD=(1)/(2)angle CAD....(2)` [ by theorem=34]
Then adding (1) and (2) we get,
`angle BDC+ angle CBD=(1)/(2) angleABC+(1)/(2) angle CAD=(1)/(2) ( angle BAC+ angle CAD)=(1)/(2) angle BAD`.
Hence `angle CBD+ angle CDB=(1)/(2) angle BAD`