\displaystyle \textbf{Question 1: }\text{In the given circle with diameter }AB,\text{ find the value of }x.\ [\text{ICSE }2003] 18d1\displaystyle \textbf{Answer:}
\displaystyle \angle ABD=\angle ACD=30^\circ\text{ (angles in the same segment)}
\displaystyle \text{Since }AB\text{ is the diameter, }\angle ADB=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle ADB,
\displaystyle x+\angle ABD+\angle ADB=180^\circ
\displaystyle x+30^\circ+90^\circ=180^\circ
\displaystyle \therefore x=60^\circ
\\

\displaystyle \textbf{Question 2: }\text{In the given figure, }O\text{ is the centre of the circle of radius }5\text{ cm. }
\displaystyle  OP\perp AB\text{ and }OQ\perp CD. \ \text{If }AB=8\text{ cm and }CD=6\text{ cm, find the length of }PQ. 18d2\displaystyle \textbf{Answer:}
\displaystyle \text{Given }OA=OC=5\text{ cm},\ OP\perp AB,\text{ and }OQ\perp CD.
\displaystyle \text{Since the perpendicular from the centre to a chord bisects the chord,}
\displaystyle AP=PB=4\text{ cm and }CQ=QD=3\text{ cm}.
\displaystyle \text{In }\triangle OAP, d1
\displaystyle OA^2=OP^2+AP^2
\displaystyle 5^2=OP^2+4^2
\displaystyle 25=OP^2+16
\displaystyle \therefore OP=3\text{ cm}
\displaystyle \text{Similarly, in }\triangle OCQ,
\displaystyle OC^2=OQ^2+CQ^2
\displaystyle 5^2=OQ^2+3^2
\displaystyle 25=OQ^2+9
\displaystyle \therefore OQ=4\text{ cm}
\displaystyle PQ=OP+OQ=3+4=7\text{ cm}
\\

\displaystyle \textbf{Question 3: }\text{The given figure shows two circles with centres }A\text{ and }
\displaystyle B\text{ and radii }5\text{ cm and }3\text{ cm, respectively, touching each other internally. }
\displaystyle \text{The perpendicular bisector of }AB\text{ meets the larger circle at }P\text{ and }Q.\text{ Find }PQ. 18d3\displaystyle \textbf{Answer:}
\displaystyle \text{Since the two circles touch internally,}
\displaystyle AB=5-3=2\text{ cm}
\displaystyle \text{Let the perpendicular bisector of }AB\text{ meet }AB\text{ at }R.
\displaystyle \therefore AR=RB=\frac{AB}{2}=1\text{ cm} d2
\displaystyle \text{Since }PQ\perp AB,\text{ we have }\angle ARP=90^\circ.
\displaystyle \text{In }\triangle APR,
\displaystyle AP^2=AR^2+PR^2
\displaystyle 5^2=1^2+PR^2
\displaystyle PR^2=25-1=24
\displaystyle \therefore PR=2\sqrt6\text{ cm}
\displaystyle \text{The perpendicular from the centre }A\text{ to chord }PQ\text{ bisects the chord.}
\displaystyle \therefore PR=RQ=2\sqrt6\text{ cm}
\displaystyle PQ=PR+RQ=2\sqrt6+2\sqrt6=4\sqrt6\text{ cm}
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\displaystyle \textbf{Question 4: }\text{In the given figure, }\triangle ABC\text{ is inscribed in a circle with centre }O\text{ and }
\displaystyle \angle BAC=30^\circ.  \  \text{Show that }BC\text{ is equal to the radius of the circumcircle of }\triangle ABC. 18d4\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\angle BAC=30^\circ.
\displaystyle \angle BOC=2\angle BAC\text{ (angle at the centre is twice the angle at the circumference subtended by the same chord)}
\displaystyle \therefore \angle BOC=2\times30^\circ=60^\circ
\displaystyle \text{In }\triangle BOC,\ OB=OC\text{ (radii of the same circle).}
\displaystyle \therefore \angle OBC=\angle OCB
\displaystyle \angle OBC+\angle OCB+\angle BOC=180^\circ
\displaystyle 2\angle OBC+60^\circ=180^\circ
\displaystyle \therefore \angle OBC=60^\circ
\displaystyle \therefore \angle OBC=\angle OCB=\angle BOC=60^\circ
\displaystyle \therefore \triangle BOC\text{ is equilateral.}
\displaystyle \therefore BC=OB=OC
\displaystyle \text{Hence, }BC\text{ is equal to the radius of the circumcircle.}
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\displaystyle \textbf{Question 5: }\text{Prove that the circle drawn on any one of the equal sides of an isosceles} \\ \text{triangle as diameter bisects the base.}
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }\triangle ABC\text{ be an isosceles triangle such that }AB=AC.
\displaystyle \text{Let the circle drawn on }AB\text{ as diameter meet the base }BC\text{ at }D.
\displaystyle \angle ADB=90^\circ\text{ (angle in a semicircle)} d4
\displaystyle \text{Since }B,D,C\text{ are collinear,}
\displaystyle \angle ADC=180^\circ-\angle ADB=90^\circ
\displaystyle \text{In }\triangle ADB\text{ and }\triangle ADC,
\displaystyle \angle ADB=\angle ADC=90^\circ
\displaystyle AB=AC\text{ (given)}
\displaystyle AD=AD\text{ (common side)}
\displaystyle \therefore \triangle ADB\cong\triangle ADC\text{ (RHS criterion)}
\displaystyle \therefore BD=DC\text{ (corresponding parts of congruent triangles)}
\displaystyle \therefore D\text{ is the midpoint of }BC.
\displaystyle \text{Hence, the circle bisects the base of the isosceles triangle.}
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\displaystyle \textbf{Question 6: }\text{In the given figure, chord }ED\text{ is parallel to diameter }
\displaystyle AC\text{ of the circle. If }\angle CBE=65^\circ,  \ \text{calculate }\angle DEC. 18d5\displaystyle \textbf{Answer:}
\displaystyle \text{Given }ED\parallel AC\text{ and }\angle CBE=65^\circ.
\displaystyle \angle EOC=2\angle EBC\text{ (angle at the centre is twice the angle at the circumference subtended by the same arc)}
\displaystyle \therefore \angle EOC=2\times65^\circ=130^\circ
\displaystyle \text{In }\triangle EOC,\ EO=OC\text{ (radii of the same circle).}
\displaystyle \therefore \angle OEC=\angle OCE
\displaystyle \angle OEC+\angle OCE+\angle EOC=180^\circ
\displaystyle 2\angle OCE+130^\circ=180^\circ
\displaystyle \therefore \angle OCE=25^\circ
\displaystyle \text{Since }A,O,C\text{ are collinear, }\angle ACE=\angle OCE=25^\circ.
\displaystyle \text{Since }ED\parallel AC,\ \angle DEC=\angle ACE\text{ (alternate interior angles).}
\displaystyle \therefore \angle DEC=25^\circ
\\

\displaystyle \textbf{Question 7: }\text{Chords }AB\text{ and }CD\text{ of a circle intersect each other at } \\ P\text{ such that }AP=CP.\text{ Prove that }AB=CD. 18d6\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AP=CP.
\displaystyle \text{By the intersecting chords theorem,}
\displaystyle AP\times PB=CP\times PD
\displaystyle \text{Since }AP=CP,
\displaystyle \therefore PB=PD
\displaystyle AB=AP+PB
\displaystyle CD=CP+PD
\displaystyle \text{Since }AP=CP\text{ and }PB=PD,
\displaystyle \therefore AP+PB=CP+PD
\displaystyle \therefore AB=CD
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\displaystyle \textbf{Question 8: }\text{Prove that the quadrilateral formed by the intersections of the angle bisectors} \\ \text{of a cyclic quadrilateral is also cyclic.}
\displaystyle \textbf{Answer:} d6\displaystyle \text{Let }ABCD\text{ be a cyclic quadrilateral.}
\displaystyle \text{Let the angle bisectors of }\angle A\text{ and }\angle D\text{ meet at }P,\text{ and those of }\angle B\text{ and }\angle C\text{ meet at }Q.
\displaystyle \text{The four angle bisectors form quadrilateral }PQRS.
\displaystyle \text{In }\triangle APD,
\displaystyle \angle PAD+\angle ADP+\angle APD=180^\circ
\displaystyle \text{Since }AP\text{ and }DP\text{ bisect }\angle DAB\text{ and }\angle ADC,\text{ respectively,}
\displaystyle \frac{1}{2}\angle DAB+\frac{1}{2}\angle ADC+\angle APD=180^\circ \qquad\text{(i)}
\displaystyle \text{In }\triangle BQC,
\displaystyle \angle QBC+\angle BCQ+\angle BQC=180^\circ
\displaystyle \text{Since }BQ\text{ and }CQ\text{ bisect }\angle ABC\text{ and }\angle BCD,\text{ respectively,}
\displaystyle \frac{1}{2}\angle ABC+\frac{1}{2}\angle BCD+\angle BQC=180^\circ \qquad\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle \frac{1}{2}(\angle DAB+\angle ABC+\angle BCD+\angle ADC)+\angle APD+\angle BQC=360^\circ
\displaystyle \text{The sum of the interior angles of }ABCD\text{ is }360^\circ.
\displaystyle \therefore 180^\circ+\angle APD+\angle BQC=360^\circ
\displaystyle \therefore \angle APD+\angle BQC=180^\circ
\displaystyle \angle APD\text{ and }\angle BQC\text{ are opposite angles of quadrilateral }PQRS.
\displaystyle \therefore PQRS\text{ is cyclic, since its opposite angles are supplementary.}
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\displaystyle \textbf{Question 9: }\text{In the given diagram, }\angle DBC=58^\circ\text{ and }BD\text{ is a} \\ \text{diameter of the circle. Calculate:}
\displaystyle \text{(i) }\angle BDC\qquad\text{(ii) }\angle BEC\qquad\text{(iii) }\angle BAC\ [\text{ICSE }2014] 18d7\displaystyle \textbf{Answer:}
\displaystyle \text{Given }BD\text{ is a diameter and }\angle DBC=58^\circ.
\displaystyle \text{(i) Since }BD\text{ is a diameter,}
\displaystyle \angle BCD=90^\circ\text{ (angle in a semicircle)}
\displaystyle \text{In }\triangle BDC,
\displaystyle \angle DBC+\angle BCD+\angle BDC=180^\circ
\displaystyle 58^\circ+90^\circ+\angle BDC=180^\circ
\displaystyle \therefore \angle BDC=32^\circ
\displaystyle \text{(ii) }BDEC\text{ is a cyclic quadrilateral.}
\displaystyle \angle BDC+\angle BEC=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle 32^\circ+\angle BEC=180^\circ
\displaystyle \therefore \angle BEC=148^\circ
\displaystyle \text{(iii) }\angle BAC=\angle BDC\text{ (angles in the same segment subtended by chord }BC\text{)}
\displaystyle \therefore \angle BAC=32^\circ
\\

\displaystyle \textbf{Question 10: }D\text{ and }E\text{ are points on the equal sides }AB\text{ and }AC\text{ of an isosceles }
\displaystyle \triangle ABC\text{ such that } \ AD=AE.\text{ Prove that the points }B,C,E\text{ and }D\text{ are concyclic.} d8\displaystyle \textbf{Answer:}
\displaystyle \text{In isosceles }\triangle ABC,\ AB=AC.
\displaystyle \text{Also, }AD=AE\text{ (given).}
\displaystyle \therefore \frac{AD}{AB}=\frac{AE}{AC}
\displaystyle \therefore DE\parallel BC\text{ (converse of the Basic Proportionality Theorem)}
\displaystyle \text{Let }\angle ABC=\angle ACB=x\text{ (base angles of an isosceles triangle).}
\displaystyle \text{Since }DE\parallel BC,\ \angle DEA=\angle BCA=x.
\displaystyle \text{In }\triangle ADE,\ AD=AE.
\displaystyle \therefore \angle ADE=\angle DEA=x
\displaystyle \text{Since }A,D,B\text{ are collinear,}
\displaystyle \angle BDE=180^\circ-\angle ADE=180^\circ-x
\displaystyle \text{Also, since }A,E,C\text{ are collinear, }\angle BCE=\angle BCA=x.
\displaystyle \therefore \angle BDE+\angle BCE=(180^\circ-x)+x=180^\circ
\displaystyle \therefore BDEC\text{ is a cyclic quadrilateral, since a pair of opposite angles is supplementary.}
\\

\displaystyle \textbf{Question 11: }\text{In the given figure, }ABCD\text{ is a cyclic quadrilateral, }
\displaystyle AF\parallel CB,\text{ and }DA\text{ is produced to }E. \ \text{If }\angle ADC=92^\circ\text{ and } \\ \angle FAE=20^\circ,\text{ determine }\angle BCD.\text{ Give a reason in support of your answer.} 18d8\displaystyle \textbf{Answer:}
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ADC+\angle ABC=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle 92^\circ+\angle ABC=180^\circ
\displaystyle \therefore \angle ABC=88^\circ
\displaystyle \text{Since }AF\parallel CB,
\displaystyle \angle BAF=\angle ABC=88^\circ\text{ (alternate interior angles)}
\displaystyle \angle BAE=\angle BAF+\angle FAE
\displaystyle \angle BAE=88^\circ+20^\circ=108^\circ
\displaystyle \text{Since }DAE\text{ is a straight line,}
\displaystyle \angle DAB+\angle BAE=180^\circ
\displaystyle \therefore \angle DAB=180^\circ-108^\circ=72^\circ
\displaystyle \angle DAB+\angle BCD=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle \therefore \angle BCD=180^\circ-72^\circ=108^\circ
\\

\displaystyle \textbf{Question 12: }\text{Let }I\text{ be the incentre of }\triangle ABC,\text{ and let }AI\text{ produced meet}
\displaystyle  \text{the circumcircle of }\triangle ABC\text{ at }D. \ \text{If }\angle BAC=66^\circ\text{ and }\angle ABC=80^\circ,\text{ calculate:}
\displaystyle \text{(i) }\angle DBC\qquad\text{(ii) }\angle IBC\qquad\text{(iii) }\angle BIC 18d9\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\angle BAC=66^\circ\text{ and }\angle ABC=80^\circ.
\displaystyle \text{(i) Since }I\text{ is the incentre, }AI\text{ bisects }\angle BAC.
\displaystyle \angle DAC=\frac{1}{2}\angle BAC=\frac{1}{2}\times66^\circ=33^\circ
\displaystyle \angle DBC=\angle DAC\text{ (angles in the same segment subtended by chord }DC\text{)}
\displaystyle \therefore \angle DBC=33^\circ
\displaystyle \text{(ii) Since }BI\text{ bisects }\angle ABC,
\displaystyle \angle IBC=\frac{1}{2}\angle ABC=\frac{1}{2}\times80^\circ=40^\circ
\displaystyle \text{(iii) In }\triangle ABC,
\displaystyle \angle ACB=180^\circ-\angle BAC-\angle ABC
\displaystyle \angle ACB=180^\circ-66^\circ-80^\circ=34^\circ
\displaystyle \text{Since }CI\text{ bisects }\angle ACB,
\displaystyle \angle ICB=\frac{1}{2}\angle ACB=\frac{1}{2}\times34^\circ=17^\circ
\displaystyle \text{In }\triangle BIC,
\displaystyle \angle BIC+\angle IBC+\angle ICB=180^\circ
\displaystyle \angle BIC+40^\circ+17^\circ=180^\circ
\displaystyle \therefore \angle BIC=123^\circ
\\

\displaystyle \textbf{Question 13: }\text{In the given figure, }AB=AD=DC=PB\text{ and } \\ \angle DBC=x^\circ.\text{ Determine, in terms of }x:
\displaystyle \text{(i) }\angle ABD\qquad\text{(ii) }\angle APB.\text{ Hence or otherwise, prove that }AP\parallel DB. 18d10\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AB=AD=DC=PB\text{ and }\angle DBC=x^\circ.
\displaystyle \text{(i) }\angle DAC=\angle DBC=x^\circ\text{ (angles in the same segment subtended by chord }DC\text{)}
\displaystyle \text{Since }AD=DC,\ \triangle ADC\text{ is isosceles.}
\displaystyle \therefore \angle ACD=\angle DAC=x^\circ
\displaystyle \angle ABD=\angle ACD\text{ (angles in the same segment subtended by chord }AD\text{)}
\displaystyle \therefore \angle ABD=x^\circ
\displaystyle \text{(ii) In }\triangle ADC,
\displaystyle \angle ADC=180^\circ-\angle DAC-\angle ACD
\displaystyle \angle ADC=180^\circ-x^\circ-x^\circ=180^\circ-2x^\circ
\displaystyle ABCD\text{ is a cyclic quadrilateral.}
\displaystyle \angle ABC+\angle ADC=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle \angle ABC=180^\circ-(180^\circ-2x^\circ)=2x^\circ
\displaystyle \text{Since }P,B,C\text{ are collinear,}
\displaystyle \angle ABP+\angle ABC=180^\circ
\displaystyle \therefore \angle ABP=180^\circ-2x^\circ
\displaystyle \text{In }\triangle ABP,\ AB=PB.
\displaystyle \therefore \angle BAP=\angle APB
\displaystyle \angle BAP+\angle APB+\angle ABP=180^\circ
\displaystyle 2\angle APB+(180^\circ-2x^\circ)=180^\circ
\displaystyle \therefore \angle APB=x^\circ
\displaystyle \angle APB=\angle DBC=x^\circ
\displaystyle \therefore AP\parallel DB\text{ (corresponding angles are equal).}
\\

\displaystyle \textbf{Question 14: }\text{In the given figure, }ABC,\ AEQ\text{ and }CEP
\displaystyle \text{ are straight lines. Show that }\angle APE\text{ and }\angle CQE \ \text{are supplementary.} 18d11\displaystyle \textbf{Answer:}
\displaystyle \text{Since }A,B,E,P\text{ lie on the same circle, }ABEP\text{ is a cyclic quadrilateral.}
\displaystyle \therefore \angle APE+\angle ABE=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}\qquad\text{(i)}
\displaystyle \text{Similarly, }B,C,E,Q\text{ lie on the same circle.}
\displaystyle \therefore \angle CQE+\angle CBE=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}\qquad\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle \angle APE+\angle CQE+\angle ABE+\angle CBE=360^\circ
\displaystyle \text{Since }A,B,C\text{ are collinear,}
\displaystyle \angle ABE+\angle CBE=180^\circ
\displaystyle \therefore \angle APE+\angle CQE+180^\circ=360^\circ
\displaystyle \therefore \angle APE+\angle CQE=180^\circ
\displaystyle \text{Hence, }\angle APE\text{ and }\angle CQE\text{ are supplementary.}
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\displaystyle \textbf{Question 15: }\text{In the given figure, }AB\text{ is the diameter of the circle with centre }O. \\ \text{ If }\angle ADC=32^\circ,\text{ find }\angle BOC. 18d12\displaystyle \textbf{Answer:}
\displaystyle \angle AOC=2\angle ADC\text{ (angle at the centre is twice the angle at the circumference subtended by the same arc)}
\displaystyle \therefore \angle AOC=2\times32^\circ=64^\circ
\displaystyle \text{Since }AB\text{ is a diameter, }A,O,B\text{ are collinear.}
\displaystyle \therefore \angle AOB=180^\circ
\displaystyle \angle AOC+\angle COB=\angle AOB
\displaystyle 64^\circ+\angle BOC=180^\circ
\displaystyle \therefore \angle BOC=116^\circ
\\

\displaystyle \textbf{Question 16: }\text{In cyclic quadrilateral }PQRS,\ \angle PQR=135^\circ.\text{ Sides }
\displaystyle SP\text{ and }RQ\text{ produced meet at }A,  \ \text{while sides }PQ\text{ and }SR\text{ produced meet at }B. \\ \text{ If }\angle A:\angle B=2:1,\text{ find }\angle A\text{ and }\angle B. d21\displaystyle \textbf{Answer:}
\displaystyle \text{Let }\angle B=x^\circ.\text{ Then }\angle A=2x^\circ.
\displaystyle \text{Since }PQRS\text{ is a cyclic quadrilateral,}
\displaystyle \angle PQR+\angle PSR=180^\circ
\displaystyle 135^\circ+\angle PSR=180^\circ
\displaystyle \therefore \angle PSR=45^\circ
\displaystyle \text{In }\triangle PBS,
\displaystyle \angle SPB+\angle PSB+\angle PBS=180^\circ
\displaystyle \angle SPB+45^\circ+x^\circ=180^\circ
\displaystyle \therefore \angle SPB=135^\circ-x^\circ \qquad\text{(i)}
\displaystyle \text{Since }A,Q,R\text{ are collinear,}
\displaystyle \angle AQP=180^\circ-\angle PQR=180^\circ-135^\circ=45^\circ
\displaystyle \text{In }\triangle APQ,
\displaystyle \angle APQ+\angle AQP+\angle PAQ=180^\circ
\displaystyle \angle APQ+45^\circ+2x^\circ=180^\circ
\displaystyle \therefore \angle APQ=135^\circ-2x^\circ
\displaystyle \text{Since }A,P,S\text{ are collinear, }\angle APQ+\angle SPB=180^\circ.
\displaystyle \therefore \angle SPB=180^\circ-(135^\circ-2x^\circ)=45^\circ+2x^\circ \qquad\text{(ii)}
\displaystyle \text{From (i) and (ii),}
\displaystyle 135^\circ-x^\circ=45^\circ+2x^\circ
\displaystyle 3x^\circ=90^\circ
\displaystyle \therefore x=30^\circ
\displaystyle \therefore \angle B=30^\circ\text{ and }\angle A=2\times30^\circ=60^\circ.
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\displaystyle \textbf{Question 17: }\text{In the following figure, }AB\text{ is the diameter of a circle with centre }
\displaystyle O,\text{ and chord }CD\text{ is equal in length} \ \text{to the radius }OA.\text{ If }AC\text{ and }BD \\ \text{ produced meet at }P,\text{ prove that }\angle APB=60^\circ. 18d13\displaystyle \textbf{Answer:}
\displaystyle \text{Given }CD=OA.
\displaystyle \text{Since }OA=OC=OD\text{ are radii of the same circle,}
\displaystyle CD=OC=OD
\displaystyle \therefore \triangle COD\text{ is equilateral.}
\displaystyle \therefore \angle COD=60^\circ
\displaystyle \therefore m\widehat{CD}=60^\circ
\displaystyle \text{Since }AB\text{ is a diameter,}
\displaystyle m\widehat{AB}=180^\circ
\displaystyle \angle APB=\frac{1}{2}\left(m\widehat{AB}-m\widehat{CD}\right)
\displaystyle \text{(angle formed by two secants outside a circle)}
\displaystyle \angle APB=\frac{1}{2}(180^\circ-60^\circ)
\displaystyle \therefore \angle APB=60^\circ
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\displaystyle \textbf{Question 18: }\text{In the following figure, }ABCD\text{ is a cyclic quadrilateral in which }
\displaystyle AD\parallel BC.\text{ The bisector of }\angle DAB \ \text{meets }BC\text{ at }E\text{ and the circle again at }F.\text{ Prove that:}
\displaystyle \text{(i) }EF=EC\qquad\text{(ii) }BF=DF 18d14\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AD\parallel BC\text{ and }AF\text{ bisects }\angle DAB.
\displaystyle \therefore \angle BAF=\angle FAD\qquad\text{(i)}
\displaystyle \text{(i) Since }AD\parallel BC\text{ and }AF\text{ is a transversal,}
\displaystyle \angle FAD=\angle AEB\text{ (alternate interior angles)}
\displaystyle \therefore \angle BAF=\angle AEB
\displaystyle \text{Since }A,E,F\text{ and }B,E,C\text{ are straight lines,}
\displaystyle \angle AEB=\angle CEF\text{ (vertically opposite angles)}
\displaystyle \therefore \angle CEF=\angle BAF
\displaystyle \angle BAF=\angle BCF\text{ (angles in the same segment subtended by chord }BF\text{)}
\displaystyle \text{Since }B,E,C\text{ are collinear, }\angle BCF=\angle ECF.
\displaystyle \therefore \angle CEF=\angle ECF
\displaystyle \therefore EF=EC\text{ (sides opposite equal angles in }\triangle ECF\text{)}
\displaystyle \text{(ii) Since }AF\text{ bisects }\angle DAB,
\displaystyle \angle BAF=\angle FAD
\displaystyle \angle BAF\text{ subtends chord }BF,\text{ and }\angle FAD\text{ subtends chord }DF.
\displaystyle \therefore BF=DF\text{ (equal angles at the circumference subtend equal chords)}
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\displaystyle \textbf{Question 19: }ABCD\text{ is a cyclic quadrilateral. Sides }AB\text{ and }DC\text{ produced meet at }
\displaystyle E,\text{ whereas sides }BC\text{ and }AD \ \text{produced meet at }F. \\ \text{ If }\angle DCF:\angle F:\angle E=3:5:4,\text{ find the angles of cyclic quadrilateral }ABCD. d23
\displaystyle \textbf{Answer:}
\displaystyle \text{Let }\angle DCF=3x,\ \angle F=5x\text{ and }\angle E=4x.
\displaystyle \text{In }\triangle CDF,\ \angle ADC\text{ is an exterior angle.}
\displaystyle \therefore \angle ADC=\angle DCF+\angle CFD
\displaystyle \angle ADC=3x+5x=8x
\displaystyle \text{In }\triangle BCE,\ \angle ABC\text{ is an exterior angle.}
\displaystyle \therefore \angle ABC=\angle BCE+\angle BEC
\displaystyle \angle BCE=\angle DCF=3x\text{ (vertically opposite angles)}
\displaystyle \therefore \angle ABC=3x+4x=7x
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle ABC+\angle ADC=180^\circ
\displaystyle 7x+8x=180^\circ
\displaystyle 15x=180^\circ
\displaystyle \therefore x=12^\circ
\displaystyle \angle ABC=7x=7\times12^\circ=84^\circ
\displaystyle \angle ADC=8x=8\times12^\circ=96^\circ
\displaystyle \text{Since }B,C,F\text{ are collinear,}
\displaystyle \angle DCB+\angle DCF=180^\circ
\displaystyle \angle DCB=180^\circ-3x=180^\circ-36^\circ=144^\circ
\displaystyle \angle DAB+\angle DCB=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle \angle DAB=180^\circ-144^\circ=36^\circ
\displaystyle \therefore \angle A=36^\circ,\ \angle B=84^\circ,\ \angle C=144^\circ\text{ and }\angle D=96^\circ.
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\displaystyle \textbf{Question 20: }\text{The following figure shows a circle with }PR\text{ as its diameter. If }
\displaystyle PQ=7\text{ cm and }QR=3RS=6\text{ cm,} \text{find the perimeter of cyclic quadrilateral }PQRS.\ [\text{ICSE }1992] 18d15\displaystyle \textbf{Answer:}
\displaystyle \text{Given }PQ=7\text{ cm and }QR=3RS=6\text{ cm}.
\displaystyle \therefore QR=6\text{ cm and }RS=\frac{6}{3}=2\text{ cm}
\displaystyle \text{Since }PR\text{ is a diameter, }\angle PQR=90^\circ\text{ (angle in a semicircle).}
\displaystyle \text{In }\triangle PQR,
\displaystyle PR^2=PQ^2+QR^2
\displaystyle PR^2=7^2+6^2=49+36=85
\displaystyle \therefore PR=\sqrt{85}\text{ cm}
\displaystyle \text{Also, }\angle PSR=90^\circ\text{ (angle in a semicircle).}
\displaystyle \text{In }\triangle PSR,
\displaystyle PR^2=PS^2+RS^2
\displaystyle 85=PS^2+2^2
\displaystyle PS^2=85-4=81
\displaystyle \therefore PS=9\text{ cm}
\displaystyle \text{Perimeter of }PQRS=PQ+QR+RS+SP
\displaystyle =7+6+2+9=24\text{ cm}
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\displaystyle \textbf{Question 21: }\text{In the following figure, }AB\text{ is the diameter of a circle with centre }O. \\ \text{ If chords }AC=AD,\text{ prove that:}
\displaystyle \text{(i) }\widehat{BC}=\widehat{DB}\qquad\text{(ii) }AB\text{ bisects }\angle CAD.
\displaystyle \text{Further, if }\widehat{AC}=2\widehat{BC},\text{ find: (a) }\angle BAC\qquad\text{(b) }\angle ABC. 18d16\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AC=AD.
\displaystyle \text{Consider }\triangle ABC\text{ and }\triangle ABD.
\displaystyle \angle ACB=\angle ADB=90^\circ\text{ (angles in a semicircle)}
\displaystyle AB=AB\text{ (common hypotenuse)}
\displaystyle AC=AD\text{ (given)}
\displaystyle \therefore \triangle ABC\cong\triangle ABD\text{ (RHS criterion)}
\displaystyle \text{(i) }BC=BD\text{ (corresponding parts of congruent triangles)}
\displaystyle \therefore \widehat{BC}=\widehat{DB}\text{ (equal chords subtend equal arcs)}
\displaystyle \text{(ii) }\angle CAB=\angle BAD\text{ (corresponding parts of congruent triangles)}
\displaystyle \therefore AB\text{ bisects }\angle CAD.
\displaystyle \text{Further, let }\widehat{BC}=x^\circ.
\displaystyle \therefore \widehat{AC}=2x^\circ
\displaystyle \text{Since }AB\text{ is a diameter, }\widehat{AC}+\widehat{CB}=180^\circ.
\displaystyle 2x+x=180^\circ
\displaystyle \therefore x=60^\circ
\displaystyle \therefore \widehat{BC}=60^\circ\text{ and }\widehat{AC}=120^\circ.
\displaystyle \text{(a) }\angle BAC=\frac{1}{2}\widehat{BC}=\frac{1}{2}\times60^\circ=30^\circ
\displaystyle \text{(b) }\angle ABC=\frac{1}{2}\widehat{AC}=\frac{1}{2}\times120^\circ=60^\circ
\\

\displaystyle \textbf{Question 22: }\text{In cyclic quadrilateral }ABCD,\ AD=BC,\ \angle BAC=30^\circ\text{ and } \\ \angle CBD=70^\circ.\text{ Find:}
\displaystyle \text{(i) }\angle BCD\qquad\text{(ii) }\angle BCA\qquad\text{(iii) }\angle ABC\qquad\text{(iv) }\angle ADC d27\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AD=BC,\ \angle BAC=30^\circ\text{ and }\angle CBD=70^\circ.
\displaystyle \angle BDC=\angle BAC=30^\circ\text{ (angles in the same segment subtended by chord }BC\text{)}
\displaystyle \angle CAD=\angle CBD=70^\circ\text{ (angles in the same segment subtended by chord }CD\text{)}
\displaystyle \angle BAD=\angle BAC+\angle CAD
\displaystyle \angle BAD=30^\circ+70^\circ=100^\circ
\displaystyle \text{(i) Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle BAD+\angle BCD=180^\circ
\displaystyle \therefore \angle BCD=180^\circ-100^\circ=80^\circ
\displaystyle \text{Since }AD=BC,\text{ equal chords subtend equal angles at the circumference.}
\displaystyle \therefore \angle ABD=\angle BAC=30^\circ
\displaystyle \text{In }\triangle ABD,
\displaystyle \angle BDA=180^\circ-\angle BAD-\angle ABD
\displaystyle \angle BDA=180^\circ-100^\circ-30^\circ=50^\circ
\displaystyle \text{(iv) }\angle ADC=\angle ADB+\angle BDC
\displaystyle \angle ADC=50^\circ+30^\circ=80^\circ
\displaystyle \text{(iii) }\angle ABC=\angle ABD+\angle DBC
\displaystyle \angle ABC=30^\circ+70^\circ=100^\circ
\displaystyle \text{(ii) In }\triangle ABC,
\displaystyle \angle BCA=180^\circ-\angle BAC-\angle ABC
\displaystyle \angle BCA=180^\circ-30^\circ-100^\circ=50^\circ
\displaystyle \therefore \angle BCD=80^\circ,\ \angle BCA=50^\circ,\ \angle ABC=100^\circ\text{ and }\angle ADC=80^\circ.
\\

\displaystyle \textbf{Question 23: }\text{In the given figure, }\angle ACE=43^\circ\text{ and }\angle CAF=62^\circ.
\displaystyle \text{ Find the values of }a,\ b\text{ and }c. \ [\text{ICSE }2007] 18d17\displaystyle \textbf{Answer:}
\displaystyle \text{Since }A,B,C\text{ are collinear and }A,E,F\text{ are collinear,}
\displaystyle \angle BAE=\angle CAF=62^\circ
\displaystyle ABDE\text{ is a cyclic quadrilateral.}
\displaystyle \angle BAE+\angle BDE=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle 62^\circ+\angle BDE=180^\circ
\displaystyle \therefore \angle BDE=118^\circ
\displaystyle \text{In }\triangle ACE,
\displaystyle \angle CAE+\angle ACE+\angle CEA=180^\circ
\displaystyle 62^\circ+43^\circ+\angle CEA=180^\circ
\displaystyle \therefore \angle CEA=75^\circ
\displaystyle \text{Since }C,D,E\text{ are collinear and }A,E,F\text{ are collinear,}
\displaystyle \angle DEF=\angle CEA=75^\circ\text{ (vertically opposite angles)}
\displaystyle \text{Since }B,D,F\text{ are collinear,}
\displaystyle b+\angle BDE=180^\circ
\displaystyle b+118^\circ=180^\circ
\displaystyle \therefore b=62^\circ
\displaystyle \text{In }\triangle DEF,
\displaystyle b+c+\angle DEF=180^\circ
\displaystyle 62^\circ+c+75^\circ=180^\circ
\displaystyle \therefore c=43^\circ
\displaystyle \text{Since }ABDE\text{ is a cyclic quadrilateral,}
\displaystyle a+\angle AED=180^\circ
\displaystyle \angle AED=180^\circ-\angle CEA=180^\circ-75^\circ=105^\circ
\displaystyle a+105^\circ=180^\circ
\displaystyle \therefore a=75^\circ
\displaystyle \therefore a=75^\circ,\quad b=62^\circ\quad\text{and}\quad c=43^\circ.
\\

\displaystyle \textbf{Question 24: }\text{In the given figure, }AB\parallel DC,\ \angle BCE=80^\circ\text{ and } \\ \angle BAC=25^\circ.\text{ Find:}
\displaystyle \text{(i) }\angle CAD\qquad\text{(ii) }\angle CBD\qquad\text{(iii) }\angle ADC 18d18\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AB\parallel DC,\ \angle BCE=80^\circ\text{ and }\angle BAC=25^\circ.
\displaystyle \text{Since }D,C,E\text{ are collinear,}
\displaystyle \angle BCD+\angle BCE=180^\circ
\displaystyle \therefore \angle BCD=180^\circ-80^\circ=100^\circ
\displaystyle \text{Since }AB\parallel DC\text{ and }AC\text{ is a transversal,}
\displaystyle \angle BAC=\angle ACD=25^\circ\text{ (alternate interior angles)}
\displaystyle \angle BCD=\angle BCA+\angle ACD
\displaystyle 100^\circ=\angle BCA+25^\circ
\displaystyle \therefore \angle BCA=75^\circ
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle ABC=180^\circ-\angle BAC-\angle BCA
\displaystyle \angle ABC=180^\circ-25^\circ-75^\circ=80^\circ
\displaystyle \text{Since }ABCD\text{ is a cyclic quadrilateral,}
\displaystyle \angle DAB+\angle BCD=180^\circ
\displaystyle \therefore \angle DAB=180^\circ-100^\circ=80^\circ
\displaystyle \text{(i) }\angle DAB=\angle DAC+\angle CAB
\displaystyle 80^\circ=\angle DAC+25^\circ
\displaystyle \therefore \angle CAD=55^\circ
\displaystyle \text{(ii) }\angle CBD=\angle CAD\text{ (angles in the same segment subtended by chord }CD\text{)}
\displaystyle \therefore \angle CBD=55^\circ
\displaystyle \text{(iii) }\angle ABC+\angle ADC=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle \therefore \angle ADC=180^\circ-80^\circ=100^\circ
\\

\displaystyle \textbf{Question 25: }ABCD\text{ is a cyclic quadrilateral in a circle with centre }O,\text{ such that }
\displaystyle AB\text{ is a diameter and chord }CD \ \text{is equal to the radius of the circle. If }AD\text{ and } \\ BC\text{ produced meet at }P,\text{ prove that }\angle APB=60^\circ. d25\displaystyle \textbf{Answer:}
\displaystyle \text{Given }CD=OC.
\displaystyle \text{Since }OC=OD\text{ (radii of the same circle),}
\displaystyle OC=OD=CD
\displaystyle \therefore \triangle OCD\text{ is equilateral.}
\displaystyle \therefore \angle COD=60^\circ
\displaystyle \therefore m\widehat{CD}=60^\circ
\displaystyle \text{Since }AB\text{ is a diameter,}
\displaystyle m\widehat{AB}=180^\circ
\displaystyle \angle APB=\frac{1}{2}\left(m\widehat{AB}-m\widehat{CD}\right)
\displaystyle \text{(angle formed by two secants outside a circle)}
\displaystyle \angle APB=\frac{1}{2}\left(180^\circ-60^\circ\right)
\displaystyle \therefore \angle APB=60^\circ
\\

\displaystyle \textbf{Question 26: }\text{In the adjoining figure, }CP\text{ bisects }\angle ACB.\text{ Prove that }DP\text{ bisects }\angle ADB. 18d19\displaystyle \textbf{Answer:}
\displaystyle \text{Given }CP\text{ bisects }\angle ACB.
\displaystyle \therefore \angle ACP=\angle PCB
\displaystyle \angle ACP=\angle ADP\text{ (angles in the same segment subtended by chord }AP\text{)}
\displaystyle \angle PCB=\angle PDB\text{ (angles in the same segment subtended by chord }PB\text{)}
\displaystyle \therefore \angle ADP=\angle PDB
\displaystyle \therefore DP\text{ bisects }\angle ADB.
\\

\displaystyle \textbf{Question 27: }\text{In the figure shown, }AD=BC,\ \angle BAC=30^\circ\text{ and } \\ \angle CBD=70^\circ.\text{ Find:}
\displaystyle \text{(i) }\angle BCD\qquad\text{(ii) }\angle BCA\qquad\text{(iii) }\angle ABC\qquad\text{(iv) }\angle ADB 18d20\displaystyle \textbf{Answer:}
\displaystyle \angle DAC=\angle DBC=70^\circ\text{ (angles in the same segment subtended by chord }DC\text{)}
\displaystyle \angle CDB=\angle CAB=30^\circ\text{ (angles in the same segment subtended by chord }CB\text{)}
\displaystyle \text{(i) In }\triangle DCB,
\displaystyle \angle BCD=180^\circ-\angle CBD-\angle CDB
\displaystyle \angle BCD=180^\circ-70^\circ-30^\circ=80^\circ
\displaystyle \text{Since }AD=BC,\text{ equal chords subtend equal angles at the circumference.}
\displaystyle \therefore \angle ACD=\angle BAC=30^\circ
\displaystyle \text{(ii) }\angle BCD=\angle BCA+\angle ACD
\displaystyle 80^\circ=\angle BCA+30^\circ
\displaystyle \therefore \angle BCA=50^\circ
\displaystyle \text{(iii) In }\triangle ABC,
\displaystyle \angle ABC=180^\circ-\angle BAC-\angle BCA
\displaystyle \angle ABC=180^\circ-30^\circ-50^\circ=100^\circ
\displaystyle \text{(iv) Since }AD=BC,
\displaystyle \angle ABD=\angle BAC=30^\circ\text{ (equal chords subtend equal angles at the circumference)}
\displaystyle \angle BAD=\angle BAC+\angle CAD
\displaystyle \angle BAD=30^\circ+70^\circ=100^\circ
\displaystyle \text{In }\triangle ABD,
\displaystyle \angle ADB=180^\circ-\angle BAD-\angle ABD
\displaystyle \angle ADB=180^\circ-100^\circ-30^\circ=50^\circ
\displaystyle \therefore \angle BCD=80^\circ,\ \angle BCA=50^\circ,\ \angle ABC=100^\circ\text{ and }\angle ADB=50^\circ.
\\

\displaystyle \textbf{Question 28: }\text{In the figure given below, }AB\text{ and }CD\text{ are two parallel chords} 
\displaystyle \text{of a circle with centre }O.\text{ If the radius of the circle is }15\text{ cm, find the distance }
\displaystyle MN\text{ between the chords of lengths } \ 24\text{ cm and }18\text{ cm, respectively.} \ [\text{ICSE }2010] 18d21\displaystyle \textbf{Answer:}
\displaystyle \text{The perpendicular from the centre of a circle to a chord bisects the chord.}
\displaystyle \therefore AM=MB=\frac{24}{2}=12\text{ cm}
\displaystyle CN=ND=\frac{18}{2}=9\text{ cm} d24
\displaystyle \text{In right-angled }\triangle OMB,
\displaystyle OB^2=OM^2+MB^2
\displaystyle 15^2=OM^2+12^2
\displaystyle OM^2=225-144=81
\displaystyle \therefore OM=9\text{ cm}
\displaystyle \text{In right-angled }\triangle OND,
\displaystyle OD^2=ON^2+ND^2
\displaystyle 15^2=ON^2+9^2
\displaystyle ON^2=225-81=144
\displaystyle \therefore ON=12\text{ cm}
\displaystyle \text{Since the chords lie on opposite sides of the centre,}
\displaystyle MN=MO+ON=9+12=21\text{ cm}
\\

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