A set of points are said to be concyclic (or cocyclic) if they lie on a common circle. All concyclic points are the same distance from the center of the circle.
Q-11 PRQ and PSQ are two right triangles with common hypotenuse PQ. Prove that ∠QPS = ∠QRS.
Given, PQ is the common hypotenuse.
To prove, ∠QPS = ∠QRS
Construct a circumcircle around the quadrilateral PQRS. This should be possible as ∠PRQ and ∠PSQ are 90°. Therefore, these angles are in the semi-circle. Angles subtended by any other chord on the circle would be less than 90° (if subtended on major arc) or more than 90° (if subtended on minor arc). Thus there is a circle with PQ as diameter and points P, R, S, and Q line on this circle. I.e., points P, Q, R and S are concyclic.
Thus, SQ is the chord, therefore, ∠QPS = ∠QRS (Angles in the same segment of the circle)
Q-12 Prove that a cyclic parallelogram is a rectangle.
Given, CFED is a cyclic parallelogram.
To prove, CFED is rectangle.
(Opposite angles of a cyclic parallelogram) … equation (1)
Also, since he quadrilateral is a parallelogram, the opposite angles are equal, therefore
Now one of the interior angle of the parallelogram is right angled, therefore parallelogram CFED is a rectangle.