AB is a diameter of a circle. PQ is such a chord of the circle that it is neither a diameter of the circle nor a interceptor of AB. By joining the points A, P and B, Q it is found that ABQP is a quadrilateral of wich `angleBAP=angleABQ`. Prove that ABQP is a cyclic trapezium.

Explanation

Let AB be a daimeter of the circle with centre at O and PQ is such a chord of the circle that is neither a diameter of the circle nor a intercepot of AB . ABQP is quadrilateral of which `angleBAP=angleABQ` .
To prove : ABQP is a cyclic trapezium.
Proof : ABQP is a cyclic quadrilateral.
`thereforeangleBAP+angleBQP=180^(@)`..........(1)
and `angleABQ+angleAPQ=180^(@)=180^(@)`.........(2)
From (1) and (2) we get , `angleBAP+angleBQP=angleABQ+angleAPQ`
or , `angleBAP+angleBQP=angleBAP+angleAPQ" " [becauseangleABQ=angleBAP]`
`angleBQP=angleAPQ`
`therefore` from (1) we get , `angleBAP+angleAPQ=180^(@)" " [becauseangleBQP=angleAPQ]`
Thus , the sum of two adjacent angles on the same side of the transversal AP of two line segments AB and PQ is `180^(@)`
`therefore` AB and PQ are parallel to each other , i.e , two opposite sides AB and PQ of the cyclic quadrilateral ABQP are parallel to each other.
Hence ABQP is a cyclic trapezium.