proof of opposite side of cyclic …

Construction :- Draw a regular quadrilateral .. A point is made in the interior of the quad... Say p ( this point is the centre of the circle made ) We know that a point surrounds 360° around it ... _________ Now pass a line segment from one vertex to the opposite vertex passing through the point p .. Note :- the line segment is not a straight line or a diagonal.... Let the quadrilateral be named as ABCD , IN WHICH A POINT P LIES IN THE INTERIOR OF IT ... WE ALSO WNOW THAT angle at the centre is twice the pripheral angle. Thus CONSIDER ABPD AND BCDP AS QUADRILATERAL.. If angle DPA is the central angle then peripheral angle is DPA/2 Similarly angle EXTERIOR DAB MEASURES Ext.DPA/ 2 Ext. Angle DPA + angle DPA = 360° thus Angle DCB + ANGLE DAB = 1/2 ENT ANGLE DPA + 1/2 ANGLE DPA = 1/2(DPA + ENT. DPA) = 1/2 ( 360°)= 180° Thus Angle DCB+DAB= 180° Similarly ANGLE ADC + ANGLE CBA = 180° THUS WE CAN STATE THAT sum of the opposite angles of a cyclic quadrilateral is 180° ... Guys it seems lengthy but actually very easy and i am sorry as i can not state the diagram of the proof..