Q-6 PQRS is a cyclic quadrilateral whose diagonals intersect at a point O. If,, find . Further, if, find.

Solution:

In the figure on circle four points are P, Q, R and S at center O

For chord SR,

(Angles in same segment)

(Opposite angles of a cyclic quadrilateral)

In

(given)

(Angles opposite to equal sides of a triangle)

()

Also,

(Angle PRS and ORS are same angles)

Q-7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral; prove that it is a rectangle.

Solution:

Given PQRS is cyclic quadrilateral inside the circle, and its diagonals, PR and SQ are the diameters of the circle,

Note that chords of equal length subtend the same angle on the circle, in this case PR and SQ are diameters, therefore the angles they subtend on the circle will be equal and also . That is, (Angles in the semi-circle). Thus, PQRS is a rectangle as each internal angle is 90°.

Q-8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

Given that, PQRS is the trapezium with, PS and QR pair of opposite equal sides.

Construct perpendiculars PM and QN on SR from vertices P and Q respectively.

Now, in and,

(Given)

(Right angles)

(Distance between the parallel lines)

Therefore, by Right Angle Hypotenuse congruence and , by Corresponding Parts of Congruent Triangles

Also,

(sum of the co-interior angles of parallel lines)

)

Thus, PQRS is a cyclic quadrilateral as the pair of opposite angles are supplementary.