 Hi and welcome to our session. Let us just ask the following question. The question says, if diagonals of a cyclic ordinator are diameters of the circle through the vertices of the ordinator prove that it is a right angle. Before solving this question, we should first be well versed with one of the property of circles which states that angle in the semicircle is a right angle. This property means that if we have a circle in which our AB is a semicircle, then the angle subtended by this arc is of 90 degree. That is, if C is any point on the semicircle, then angle ACB is equal to 90 degree. The knowledge of this property is the key idea in this question. Now, begin with the solution. Let us first make a diagram of this question. This is the cyclic quadrilateral ABCD. Diagnose of this quadrilateral are diameters of this circle and we have to prove that ABCD is a rectangle. Let us first write down the given information. It is given to us that ABCD is a cyclic quadrilateral in which diagonals AC and BD are diameters of the circle. We have to prove that CD is a rectangle. Is the diameter, therefore, is a semicircle, right? This is a semicircle. ABC is equal to 90 degree because angle in a semicircle is a right angle and similarly when AC is the diameter then angle ADC is also equal to 90 degree because angle in a semicircle is a right angle. Now, consider diameter BD. BD is a diameter but BD is a semicircle and hence angle BAD is equal to 90 degree because angle in a semicircle is a right angle. When BD is the diameter then angle BCD is also equal to 90 degree. So, we can now say that all angles of this quadrilateral is of 90 degree and this implies that ABCD is a rectangle because in rectangle each angle is of 90 degree. Since all angles of quadrilateral ABCD is of 90 degree therefore ABCD is a rectangle because in a rectangle each angle is of 90 degree. Hence, we have proved that ABCD is a rectangle. This completes the session. Bye and take care.