NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.6 – Part 4

Angle ABC bisected by line segment BE

BE Is Angle Bisector of Angle ABC

Angle ABC bisected by line segment BE

Angle Bisector. A line that splits an angle into two equal angles. ("Bisect" means to divide into two equal parts.)

Q-7 PQ and RS are chords of a circle which bisect each other. Prove that

  1. PQ and RS are diameters

  2. PRQS is a rectangle.

Solution:

Circle c Circle c: Circle through B with center O Segment f Segment f: Segment [P, R] Segment g Segment g: Segment [R, Q] Segment h Segment h: Segment [Q, S] Segment i Segment i: Segment [P, S] Segment j Segment j: Segment [P, Q] Segment k Segment k: Segment [R, S] Point O O = (0.44, 2.76) Point O O = (0.44, 2.76) Point O O = (0.44, 2.76) 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 Point S Point S: Point on c Point S Point S: Point on c Point S Point S: Point on c

PQ and RS Bisect Each Other at Point O

PQ and RS are chords of circle which bisect each other at O.

Solution (i): To prove PQ and RS are diameters

Given,

  • PQ and RS are two chords of a circle which intersect at O.

  • They bisects each other at O

Construction,

Join PS, SQ, QR and RP.

In SOQandROP

  • OS=OR (O is the mid-point of SR)

  • SOQ=ROP (Vertically opposite angles)

  • OQ=OP (O is the mid-point of SQ)

  • Therefore, SOQROP by Side-Angle-Side criterion of congruence.

  • Therefore, PQ=SR

  • Since segment PQ and SR are equal, corresponding arcs in the circle are also equal, therefore arc(QS)=arc(PR) ………..equation (1)

Similarly, in SOP and ROQ

  • arc(PS)=arc(QR) ……….equation (2)

From equation (1) and (2)

  • arc(QS)+arc(PS)=arc(PR)+arc(QR)

  • arc(QSP)=arc(PRP)

Therefore PQ divides the circle into two parts

Therefore PQ is a diameter.

Similarly, we can prove that PR is a diameter.

Solution (ii): To prove PRQS is a rectangle.

SOQROP (Proved above), therefore, OSQ(orRPQ)=ORPorSRP

With two lines PR and SQ intersected by transversal SR the pair of interior opposite angles are equal, therefore, SQRP

Similarly, POSQOR and PSRQ

Since PSRQ and SQRP , Therefore PRQS is a cyclic parallelogram

Also, SPR=SQR ……….equation (3) (because opposite angle of a parallelogram are equal)

Since PRQS is a cyclic quadrilateral

Therefore pair of opposite angles are supplementary, therefore SPR+SQR=180° ……….equation (4)

From equation (3) and (4)

  • SQR+SQR=180° ( SPR=SQR )

  • 2SQR=180

  • SQR=SPR=90°

Since opposite angles of parallelogram PRQS are 90° each, therefore, PRQS is a rectangle.

Q-8 Bisectors of angle P, Q and R of a triangle PQR intersect its circumcircle at C, A and B respectively, Prove that the angles of the triangle CAB are 90°12A,90°12B,90°12C.

Solution:

Circle c Circle c: Circle through R with center O Segment f Segment f: Segment [P, Q] Segment g Segment g: Segment [Q, A] Segment h Segment h: Segment [P, C] Segment i Segment i: Segment [P, R] Segment j Segment j: Segment [Q, R] Segment k Segment k: Segment [A, B] Segment l Segment l: Segment [B, C] Segment m Segment m: Segment [C, A] Segment n Segment n: Segment [B, R] Point O O = (0.8, 2.7) Point O O = (0.8, 2.7) Point O O = (0.8, 2.7) Point R R = (0.92, 0.64) Point R R = (0.92, 0.64) Point R R = (0.92, 0.64) Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point A Point A: Point on c Point A Point A: Point on c Point A Point A: Point on c Point C Point C: Point on c Point C Point C: Point on c Point C Point C: Point on c Point B Point B: Point on c Point B Point B: Point on c Point B Point B: Point on c

Bisectors of Triangle PQR Intersect at Its Circumcircle

Bisectors of a triangle PQR intersect at its circumcircle centered at O

  • C=ACB=ACP+PCB

  • Since ACPandAQP are angles in same segment of circle (AP), therefore ACP=AQP . Similarly, PCB=BRP . Therefore equation above can be written as C=AQP+BRP=12Q+12R (Since QA and RB are angle bisectors of Q and R .

  • Therefore, C=Q+R2

Similarly,

  • A=P+R2 and B=(P+Q)2

Also,

  • C=180°P2=90°12P ( P+Q+R=180° )

  • A=180°Q2=90°12Q ( P+Q+R=180° )

  • B=180°R2=90°12R ( P+Q+R=180° )

Now,

  • C=90°12P

  • A=90°12Q

  • B=90°12R

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