Circles C (O,r) and C(O', r'), (r r' ) touch internally at P. PQ is a chord of circle C (O,r) which intersect circle C (O' , r' ) at R. Show that OO'RQ is a Trapezium.

 Angle included between the chord and tangent of a circle is equal to the angle in the alternate segment. Also the angle subtended by the chord at the center is twice the angle in the segment of the same chord.

∠PO'R = ∠POQ

But these two angles are corresponding angles; 
since they are equal OQ is parallel to O'P.

AS well QR & OO' are not parallel, since they are two distinct lines having a common vertex P.

Hence from the above two, a pair of opposite lines of the quadrilateral OO'RQ is parallel. 
So OO'RQ is a trapezium.