NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.5 – Part 5

Cocyclic points on a circle

Concyclic Points

Cocyclic points on a circle

  • 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.

Solution:

Circle c Circle c: Circle through S with center O Angle ? Angle ?: Angle between Q, R, E' Angle ? Angle ?: Angle between Q, R, E' Angle ? Angle ?: Angle between Q, R, E' Angle ? Angle ?: Angle between Q, S, E'_1 Angle ? Angle ?: Angle between Q, S, E'_1 Angle ? Angle ?: Angle between Q, S, E'_1 Segment f Segment f: Segment [P, R] Segment g Segment g: Segment [R, Q] Segment h Segment h: Segment [P, Q] Segment i Segment i: Segment [R, S] Segment j Segment j: Segment [Q, S] Segment k Segment k: Segment [P, S] Point O O = (-0.62, 4.12) Point O O = (-0.62, 4.12) Point O O = (-0.62, 4.12) Point S S = (-0.36, 1.98) Point S S = (-0.36, 1.98) Point S S = (-0.36, 1.98) Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c Point R Point R: Point on c Point R Point R: Point on c Point R Point R: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c

Circle and Two Right Triangles

PRQ and PSQ are two right triangles with common hypotenuse PQ

Given, PQ is the common hypotenuse. R=S=90°.

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.

Solution:

Circle c Circle c: Circle through B with center A Angle ? Angle ?: Angle between C, D, E Angle ? Angle ?: Angle between C, D, E Angle ? Angle ?: Angle between C, D, E Angle ? Angle ?: Angle between C, F, E Angle ? Angle ?: Angle between C, F, E Angle ? Angle ?: Angle between C, F, E Segment f Segment f: Segment [C, D] Segment g Segment g: Segment [D, E] Segment h Segment h: Segment [E, F] Segment i Segment i: Segment [F, C] Point C Point C: Point on c Point C Point C: Point on c Point C Point C: Point on c Point D Point D: Point on c Point D Point D: Point on c Point D Point D: Point on c Point E Point E: Point on c Point E Point E: Point on c Point E Point E: Point on c Point F Point F: Point on c Point F Point F: Point on c Point F Point F: Point on c

Parallelogram CFED Inside a Circle

Parallelogram CFED, with angles α and β

Given, CFED is a cyclic parallelogram.

To prove, CFED is rectangle.

β+α=180° (Opposite angles of a cyclic parallelogram) … equation (1)

Also, since he quadrilateral is a parallelogram, the opposite angles are equal, therefore β=α

So,

  • β+α=180°

  • β+β=180 ( β=α )

  • 2β=180

  • β=90

Now one of the interior angle of the parallelogram is right angled, therefore parallelogram CFED is a rectangle.

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