\displaystyle \textbf{Question 1: }\text{In the given diagram, chords }AB\text{ and }BC\text{ are equal.}
\displaystyle \text{(i) What is the relation between }\widehat{AB}\text{ and }\widehat{BC}\text{?}
\displaystyle \text{(ii) What is the relation between }\angle AOB\text{ and }\angle BOC\text{?}
\displaystyle \text{(iii) If the minor arc }\widehat{AD}\text{ is greater than the minor arc }\widehat{ABC},\text{ compare chords }AD\text{ and }AC.
\displaystyle \text{(iv) If }\angle AOB=50^\circ,\text{ find the measure of }\angle BAC. 18c10\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AB=BC.
\displaystyle \text{(i) }\widehat{AB}=\widehat{BC}
\displaystyle \text{Equal chords of a circle subtend equal arcs.} c1
\displaystyle \text{(ii) }\angle AOB=\angle BOC
\displaystyle \text{Equal chords of a circle subtend equal angles at the centre.}
\displaystyle \text{(iii) Since }\widehat{AD}>\widehat{ABC},
\displaystyle \therefore AD>AC
\displaystyle \text{The greater minor arc subtends the greater chord.}
\displaystyle \text{(iv) Since }AB=BC,\text{ we have }\angle AOB=\angle BOC.
\displaystyle \therefore \angle BOC=50^\circ
\displaystyle \angle BAC=\frac{1}{2}\angle BOC
\displaystyle \therefore \angle BAC=\frac{1}{2}\times 50^\circ=25^\circ
\\

\displaystyle \textbf{Question 2: }\text{In }\triangle ABC,\text{ the perpendiculars drawn from vertices }A\text{ and }B\text{ to their opposite sides}
\displaystyle \text{meet the circumcircle of }\triangle ABC\text{ again at }D\text{ and } \\ E\text{ respectively. Prove that }\widehat{CD}=\widehat{CE}. c2.jpg\displaystyle \textbf{Answer:}
\displaystyle \text{Since }AD\perp BC,\text{ we have }\angle ADC=90^\circ.
\displaystyle \therefore \angle CAD=90^\circ-\angle ACD
\displaystyle \text{Since }BE\perp AC,\text{ we have }\angle BEC=90^\circ.
\displaystyle \therefore \angle CBE=90^\circ-\angle BCE
\displaystyle \text{Since }A,B,C,D\text{ and }E\text{ lie on the same circle,}
\displaystyle \angle ACD=\angle ABD\text{ and }\angle BCE=\angle BAE.
\displaystyle \text{However, a more direct comparison gives}
\displaystyle \angle CAD=90^\circ-\angle ACB
\displaystyle \angle CBE=90^\circ-\angle BCA
\displaystyle \therefore \angle CAD=\angle CBE
\displaystyle \text{Equal angles at the circumference subtend equal arcs.}
\displaystyle \therefore \widehat{CD}=\widehat{CE}
\\

\displaystyle \textbf{Question 3: }\text{In a cyclic trapezium, prove that the non-parallel sides are equal and the} \\ \text{diagonals are also equal.}
\displaystyle \textbf{Answer:} c3
\displaystyle \text{Let }ABCD\text{ be a cyclic trapezium such that }AB\parallel DC.
\displaystyle \angle ABD=\angle BDC\text{ (alternate interior angles)}
\displaystyle \angle ABD\text{ subtends chord }AD\text{ and }\angle BDC\text{ subtends chord }BC.
\displaystyle \therefore AD=BC\text{ (equal angles at the circumference subtend equal chords)}
\displaystyle \therefore \widehat{AD}=\widehat{BC}\text{ (equal chords subtend equal arcs)}
\displaystyle \therefore \widehat{ADC}=\widehat{BCD}\text{ (adding the common arc }\widehat{DC}\text{)}
\displaystyle \therefore AC=BD\text{ (equal arcs subtend equal chords)}
\\

\displaystyle \textbf{Question 4: }\text{In the given diagram, }AD\text{ is the diameter of the circle with centre }
\displaystyle O.\text{ Chords }AB=BC=CD.  \ \text{If }\angle DEF=110^\circ,\text{ find (i) }\angle AEF\text{ and (ii) }\angle FAB. 18c9\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AB=BC=CD.
\displaystyle \text{(i) Since }AD\text{ is a diameter,}
\displaystyle \angle AED=90^\circ\text{ (angle in a semicircle)}
\displaystyle \angle DEF=\angle DEA+\angle AEF
\displaystyle 110^\circ=90^\circ+\angle AEF
\displaystyle \therefore \angle AEF=20^\circ
\displaystyle \text{(ii) Since }AB=BC=CD,
\displaystyle \angle AOB=\angle BOC=\angle COD\text{ (equal chords subtend equal angles at the centre)}
\displaystyle \angle AOB+\angle BOC+\angle COD=\angle AOD=180^\circ
\displaystyle \therefore \angle AOB=\angle BOC=\angle COD=60^\circ
\displaystyle \text{In }\triangle AOB,\ OA=OB\text{ (radii of the same circle).}
\displaystyle \therefore \angle OAB=\angle OBA
\displaystyle \angle OAB=\frac{180^\circ-\angle AOB}{2}=\frac{180^\circ-60^\circ}{2}=60^\circ
\displaystyle ADEF\text{ is a cyclic quadrilateral.}
\displaystyle \angle DAF+\angle DEF=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle \therefore \angle DAF=180^\circ-110^\circ=70^\circ
\displaystyle \angle FAB=\angle FAD+\angle DAB
\displaystyle \angle FAB=70^\circ+60^\circ=130^\circ
\\

\displaystyle \textbf{Question 5: }\text{In the given diagram, if }\widehat{AB}=\widehat{CD}\text{ in a circle with centre }O,
\displaystyle \text{ prove that quadrilateral }ABCD \ \text{is an isosceles trapezium.} 18c8\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\widehat{AB}=\widehat{CD}.
\displaystyle \angle ADB=\angle DBC\text{ (equal arcs subtend equal angles at the circumference)}
\displaystyle \therefore AD\parallel BC\text{ (alternate interior angles are equal)}
\displaystyle \therefore ABCD\text{ is a trapezium.}
\displaystyle \text{Also, }\widehat{AB}=\widehat{CD}.
\displaystyle \therefore AB=CD\text{ (equal arcs subtend equal chords)}
\displaystyle \text{Thus, the non-parallel sides }AB\text{ and }CD\text{ are equal.}
\displaystyle \therefore ABCD\text{ is an isosceles trapezium.}
\\

\displaystyle \textbf{Question 6: }\text{In the given figure, }\triangle ABC\text{ is an isosceles triangle with } \\ AB=AC,\text{ and }O\text{ is the centre of its circumcircle.}
\displaystyle \text{Prove that }AP\text{ bisects }\angle BPC. 18c7\displaystyle \textbf{Answer:}
\displaystyle \text{Since }\triangle ABC\text{ is isosceles, }AB=AC.
\displaystyle \angle APB\text{ subtends chord }AB,\text{ and }\angle APC\text{ subtends chord }AC.
\displaystyle \therefore \angle APB=\angle APC\text{ (equal chords subtend equal angles at the circumference)}
\displaystyle \therefore AP\text{ bisects }\angle BPC.
\\

\displaystyle \textbf{Question 7: }\text{If two sides of a cyclic quadrilateral are parallel, prove that:}
\displaystyle \text{(i) its other two sides are equal}
\displaystyle \text{(ii) its diagonals are equal.}
\displaystyle \textbf{Answer:} c7
\displaystyle \text{Let }ABCD\text{ be a cyclic quadrilateral such that }AB\parallel DC.
\displaystyle \text{(i) Since }AB\parallel DC,
\displaystyle \angle BAC=\angle DCA\text{ (alternate interior angles)}
\displaystyle \angle BAC\text{ subtends chord }BC,\text{ and }\angle DCA\text{ subtends chord }AD.
\displaystyle \therefore AD=BC\text{ (equal angles at the circumference subtend equal chords)}
\displaystyle \text{Hence, the other two sides are equal.}
\displaystyle \text{(ii) Since }AD=BC,
\displaystyle \widehat{AD}=\widehat{BC}\text{ (equal chords subtend equal arcs)}
\displaystyle \text{Adding the common arc }\widehat{AB}\text{ to both sides,}
\displaystyle \widehat{AD}+\widehat{AB}=\widehat{BC}+\widehat{AB}
\displaystyle \therefore \widehat{DAB}=\widehat{ABC}
\displaystyle \therefore DB=AC\text{ (equal arcs subtend equal chords)}
\displaystyle \text{Hence, the diagonals are equal.}
\\

\displaystyle \textbf{Question 8: }\text{In the given diagram, the circle has centre }O.\text{ If }PQ=QR=RS\text{ and } \\ \angle PTS=75^\circ,\text{ calculate:}
\displaystyle \text{(i) }\angle POS
\displaystyle \text{(ii) }\angle QOR
\displaystyle \text{(iii) }\angle PQR 18c6\displaystyle \textbf{Answer:}
\displaystyle \text{Given }PQ=QR=RS\text{ and }\angle PTS=75^\circ.
\displaystyle \text{(i) }\angle POS=2\angle PTS
\displaystyle \angle POS=2\times75^\circ=150^\circ
\displaystyle \text{Therefore, }\angle POS=150^\circ.
\displaystyle \text{(ii) Since }PQ=QR=RS,
\displaystyle \angle POQ=\angle QOR=\angle ROS\text{ (equal chords subtend equal angles at the centre)}
\displaystyle \angle POQ+\angle QOR+\angle ROS=\angle POS=150^\circ
\displaystyle \therefore \angle QOR=\frac{150^\circ}{3}=50^\circ
\displaystyle \text{Therefore, }\angle QOR=50^\circ.
\displaystyle \text{(iii) In }\triangle OPQ,\ OP=OQ\text{ (radii of the same circle).}
\displaystyle \therefore \angle OPQ=\angle OQP
\displaystyle \angle OPQ+\angle OQP=180^\circ-\angle POQ
\displaystyle \angle OPQ+\angle OQP=180^\circ-50^\circ=130^\circ
\displaystyle \therefore \angle OQP=65^\circ
\displaystyle \text{Similarly, in }\triangle OQR,\ \angle OQR=65^\circ.
\displaystyle \angle PQR=\angle PQO+\angle OQR
\displaystyle \angle PQR=65^\circ+65^\circ=130^\circ
\displaystyle \text{Therefore, }\angle PQR=130^\circ.
\\

\displaystyle \textbf{Question 9: }\text{In the given diagram, }AB\text{ is a side of a regular hexagon and }AC\text{ is a side of}
\displaystyle \text{a regular octagon inscribed in a circle with centre }O.\text{ Calculate:}
\displaystyle \text{(i) }\angle AOB
\displaystyle \text{(ii) }\angle ACB
\displaystyle \text{(iii) }\angle ABC 18c5\displaystyle \textbf{Answer:}
\displaystyle \text{(i) Since }AB\text{ is a side of a regular hexagon,}
\displaystyle \angle AOB=\frac{360^\circ}{6}=60^\circ
\displaystyle \text{(ii) }\angle ACB\text{ and }\angle AOB\text{ are subtended by the same arc }\widehat{AB}.
\displaystyle \therefore \angle ACB=\frac{1}{2}\angle AOB
\displaystyle \angle ACB=\frac{1}{2}\times60^\circ=30^\circ
\displaystyle \text{(iii) Since }AC\text{ is a side of a regular octagon,}
\displaystyle \angle AOC=\frac{360^\circ}{8}=45^\circ
\displaystyle \angle ABC\text{ and }\angle AOC\text{ are subtended by the same arc }\widehat{AC}.
\displaystyle \therefore \angle ABC=\frac{1}{2}\angle AOC
\displaystyle \angle ABC=\frac{1}{2}\times45^\circ=22.5^\circ
\\

\displaystyle \textbf{Question 10: }\text{In a regular pentagon }ABCDE\text{ inscribed in a circle, find the ratio } \\ \angle EDA:\angle ADC.\ [\text{ICSE } 1990] c10.jpg\displaystyle \textbf{Answer:}
\displaystyle \text{Since }ABCDE\text{ is a regular pentagon, each side subtends an arc of }
\displaystyle \frac{360^\circ}{5}=72^\circ.
\displaystyle \angle EDA\text{ is subtended by arc }\widehat{EA}.
\displaystyle \therefore \angle EDA=\frac{1}{2}\times72^\circ=36^\circ
\displaystyle \angle ADC\text{ is subtended by arc }\widehat{ABC}.
\displaystyle \widehat{ABC}=\widehat{AB}+\widehat{BC}=72^\circ+72^\circ=144^\circ
\displaystyle \therefore \angle ADC=\frac{1}{2}\times144^\circ=72^\circ
\displaystyle \therefore \angle EDA:\angle ADC=36^\circ:72^\circ=1:2
\\

\displaystyle \textbf{Question 11: }\text{In the given diagram, }AB=BC=CD\text{ and }\angle ABC=132^\circ.\text{ Find:}
\displaystyle \text{(i) }\angle AEB
\displaystyle \text{(ii) }\angle AED
\displaystyle \text{(iii) }\angle COD\ [\text{ICSE } 1993] 18c4\displaystyle \textbf{Answer:}
\displaystyle \text{Given }AB=BC=CD\text{ and }\angle ABC=132^\circ.
\displaystyle \text{(i) }ABCE\text{ is a cyclic quadrilateral.}
\displaystyle \angle ABC+\angle AEC=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle 132^\circ+\angle AEC=180^\circ
\displaystyle \therefore \angle AEC=48^\circ
\displaystyle \text{Since }AB=BC,\ \angle AEB=\angle BEC\text{ (equal chords subtend equal angles at the circumference)}
\displaystyle \angle AEB+\angle BEC=\angle AEC=48^\circ
\displaystyle \therefore 2\angle AEB=48^\circ
\displaystyle \therefore \angle AEB=24^\circ
\displaystyle \text{(ii) Since }AB=BC=CD,
\displaystyle \angle AEB=\angle BEC=\angle CED=24^\circ
\displaystyle \angle AED=\angle AEB+\angle BEC+\angle CED
\displaystyle \therefore \angle AED=24^\circ+24^\circ+24^\circ=72^\circ
\displaystyle \text{(iii) }\angle COD\text{ and }\angle CED\text{ are subtended by the same arc }\widehat{CD}.
\displaystyle \angle COD=2\angle CED
\displaystyle \therefore \angle COD=2\times24^\circ=48^\circ
\\

\displaystyle \textbf{Question 12: }\text{In the diagram, }O\text{ is the centre of the circle and }\widehat{AB}=2\widehat{BC}. \\ \text{ If }\angle AOB=108^\circ,\text{ find:}
\displaystyle \text{(i) }\angle CAB
\displaystyle \text{(ii) }\angle ADB\ [\text{ICSE } 1996] c12.jpg\displaystyle \textbf{Answer:}
\displaystyle \text{Given }\widehat{AB}=2\widehat{BC}\text{ and }\angle AOB=108^\circ.
\displaystyle \text{Since the angles at the centre are proportional to their corresponding arcs,}
\displaystyle \angle AOB=2\angle BOC
\displaystyle \therefore \angle BOC=\frac{108^\circ}{2}=54^\circ
\displaystyle \text{(i) }\angle CAB\text{ and }\angle COB\text{ are subtended by the same arc }\widehat{CB}.
\displaystyle \angle CAB=\frac{1}{2}\angle COB
\displaystyle \therefore \angle CAB=\frac{1}{2}\times54^\circ=27^\circ
\displaystyle \text{(ii) }\angle ACB\text{ and }\angle AOB\text{ are subtended by the same arc }\widehat{AB}.
\displaystyle \angle ACB=\frac{1}{2}\angle AOB
\displaystyle \therefore \angle ACB=\frac{1}{2}\times108^\circ=54^\circ
\displaystyle ADBC\text{ is a cyclic quadrilateral.}
\displaystyle \angle ADB+\angle ACB=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle \angle ADB+54^\circ=180^\circ
\displaystyle \therefore \angle ADB=126^\circ
\\

\displaystyle \textbf{Question 13: }\text{In the diagram given below, }O\text{ is the centre of the circle, }AB\text{ is a side of }
\displaystyle \text{a regular pentagon, and AC is a side of a regular hexagon. Find the angles of }\triangle ABC. c13\displaystyle \textbf{Answer:}
\displaystyle \text{Since }AB\text{ is a side of a regular pentagon,}
\displaystyle \angle AOB=\frac{360^\circ}{5}=72^\circ
\displaystyle \text{Since }AC\text{ is a side of a regular hexagon,}
\displaystyle \angle AOC=\frac{360^\circ}{6}=60^\circ
\displaystyle \text{Therefore, reflex }\angle BOC=360^\circ-72^\circ-60^\circ=228^\circ
\displaystyle \angle BAC\text{ is subtended by the major arc }\widehat{BC}.
\displaystyle \therefore \angle BAC=\frac{1}{2}\times228^\circ=114^\circ
\displaystyle \angle ABC\text{ and }\angle AOC\text{ are subtended by the same arc }\widehat{AC}.
\displaystyle \therefore \angle ABC=\frac{1}{2}\times60^\circ=30^\circ
\displaystyle \angle ACB\text{ and }\angle AOB\text{ are subtended by the same arc }\widehat{AB}.
\displaystyle \therefore \angle ACB=\frac{1}{2}\times72^\circ=36^\circ
\displaystyle \therefore \angle ABC=30^\circ,\ \angle ACB=36^\circ\text{ and }\angle BAC=114^\circ.
\\

\displaystyle \textbf{Question 14: }\text{In the given diagram, }BD\text{ is a side of a regular hexagon, }DC
\displaystyle \text{ is a side of a regular pentagon, and AD}  \ \text{is a diameter. Calculate:}
\displaystyle \text{(i) }\angle ADC
\displaystyle \text{(ii) }\angle BDA
\displaystyle \text{(iii) }\angle ABC
\displaystyle \text{(iv) }\angle AEC\ [\text{ICSE } 1984] 18c1\displaystyle \textbf{Answer:}
\displaystyle \text{Since }BD\text{ is a side of a regular hexagon,}
\displaystyle \angle BOD=\frac{360^\circ}{6}=60^\circ
\displaystyle \text{Since }DC\text{ is a side of a regular pentagon,}
\displaystyle \angle DOC=\frac{360^\circ}{5}=72^\circ
\displaystyle \text{(i) In }\triangle OCD,\ OC=OD\text{ (radii of the same circle).}
\displaystyle \therefore \angle ODC=\angle OCD=\frac{180^\circ-72^\circ}{2}=54^\circ
\displaystyle \text{Since }A,O,D\text{ are collinear, }\angle ADC=\angle ODC.
\displaystyle \therefore \angle ADC=54^\circ
\displaystyle \text{(ii) In }\triangle BOD,\ OB=OD\text{ (radii of the same circle).}
\displaystyle \angle BOD=60^\circ
\displaystyle \therefore \angle BDO=\angle OBD=\frac{180^\circ-60^\circ}{2}=60^\circ
\displaystyle \text{Since }A,O,D\text{ are collinear, }\angle BDA=\angle BDO.
\displaystyle \therefore \angle BDA=60^\circ
\displaystyle \text{(iii) }\angle ABC=\angle ADC\text{ (angles in the same segment subtended by chord }AC\text{)}
\displaystyle \therefore \angle ABC=54^\circ
\displaystyle \text{(iv) }AECD\text{ is a cyclic quadrilateral.}
\displaystyle \angle AEC+\angle ADC=180^\circ\text{ (opposite angles of a cyclic quadrilateral)}
\displaystyle \therefore \angle AEC=180^\circ-54^\circ=126^\circ
\\

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