\displaystyle \textbf{Question 1: }\text{In an equilateral triangle, prove that the centroid and the center of the}
\displaystyle \text{circum-circle (circumcenter) coincide.}
\displaystyle \text{Answer:} 2018-12-02_18-36-06\displaystyle \text{Given: }\triangle ABC\text{ is an equilateral triangle. }D,E\text{ and }F\text{ are the midpoints}
\displaystyle \text{of }BC,CA\text{ and }AB,\text{ respectively.}
\displaystyle \text{To prove: The centroid and the circumcenter of }\triangle ABC\text{ coincide.}
\displaystyle \text{Construction: Draw the medians }AD,BE\text{ and }CF,\text{ intersecting at }O.
\displaystyle \text{Proof: In }\triangle ABD\text{ and }\triangle ACD,
\displaystyle AB=AC\qquad\text{[Sides of an equilateral triangle]}
\displaystyle BD=DC\qquad\text{[Since }D\text{ is the midpoint of }BC\text{]}
\displaystyle AD=AD\qquad\text{[Common]}
\displaystyle \therefore \triangle ABD\cong\triangle ACD\qquad\text{[By S.S.S. criterion]}
\displaystyle \therefore \angle ADB=\angle ADC\qquad\text{[By C.P.C.T.C.]}
\displaystyle \text{But }\angle ADB+\angle ADC=180^\circ.
\displaystyle \therefore \angle ADB=\angle ADC=90^\circ
\displaystyle \therefore AD\perp BC.
\displaystyle \text{Since }D\text{ is the midpoint of }BC,\ AD\text{ is the perpendicular bisector of }BC.
\displaystyle \text{Similarly, }BE\text{ and }CF\text{ are the perpendicular bisectors of }CA\text{ and }AB.
\displaystyle \text{Therefore, }O,\text{ the point of intersection of the medians, is the centroid of }\triangle ABC.
\displaystyle \text{Also, }O\text{ is the point of intersection of the perpendicular bisectors of its sides.}
\displaystyle \therefore O\text{ is the circumcenter of }\triangle ABC.
\displaystyle \therefore \text{The centroid and the circumcenter of an equilateral triangle coincide.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Two circles whose centers are }O\text{ and }O'\text{ intersect at }P.\text{ Through }P,
\displaystyle \text{a line }l\parallel OO'\text{ intersecting the circles at }C\text{ and }D\text{ is drawn. Prove that}
\displaystyle CD=2OO'.
\displaystyle \text{Answer:} 2018-12-02_18-35-53\displaystyle \text{Given: Two circles with centers }O\text{ and }O'\text{ intersect at }P,\text{ and }CD\parallel OO'.
\displaystyle \text{To prove: }CD=2OO'.
\displaystyle \text{Construction: Draw }OA\perp CD\text{ and }O'B\perp CD.
\displaystyle \text{Proof: Since }OA\perp CD\text{ and }CP\text{ is a chord of the circle with center }O,
\displaystyle \text{the perpendicular from the center to a chord bisects the chord.}
\displaystyle \therefore CA=AP
\displaystyle \therefore CP=CA+AP=2AP\qquad\ldots\text{(i)}
\displaystyle \text{Similarly, since }O'B\perp CD\text{ and }PD\text{ is a chord of the circle with center }O',
\displaystyle PB=BD
\displaystyle \therefore PD=PB+BD=2PB\qquad\ldots\text{(ii)}
\displaystyle \text{Now, }CD=CP+PD
\displaystyle =2AP+2PB
\displaystyle =2(AP+PB)
\displaystyle =2AB\qquad\ldots\text{(iii)}
\displaystyle \text{Since }CD\parallel OO',\ AB\parallel OO'.
\displaystyle \text{Also, }OA\perp AB\text{ and }O'B\perp AB.
\displaystyle \therefore OA\parallel O'B.
\displaystyle \therefore OABO'\text{ is a rectangle.}
\displaystyle \therefore AB=OO'\qquad\text{[Opposite sides of a rectangle]}
\displaystyle \text{From (iii), }CD=2AB=2OO'.
\displaystyle \therefore CD=2OO'.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Prove that the line joining the midpoints of two parallel chords of a circle}
\displaystyle \text{passes through the center.}
\displaystyle \text{Answer:} 2018-12-02_18-35-21\displaystyle \text{Given: }AB\text{ and }CD\text{ are two parallel chords of a circle with center }O.
\displaystyle P\text{ and }Q\text{ are the midpoints of }AB\text{ and }CD,\text{ respectively.}
\displaystyle \text{To prove: The line }PQ\text{ passes through the center }O.
\displaystyle \text{Construction: Join }OP\text{ and }OQ.
\displaystyle \text{Proof: Since }P\text{ is the midpoint of chord }AB,
\displaystyle OP\perp AB
\displaystyle \text{[The line joining the center to the midpoint of a chord is perpendicular to the chord]}
\displaystyle \text{Similarly, since }Q\text{ is the midpoint of chord }CD,
\displaystyle OQ\perp CD
\displaystyle \text{But }AB\parallel CD.
\displaystyle \therefore OP\parallel OQ
\displaystyle \text{Since }OP\text{ and }OQ\text{ both pass through the point }O,\text{ they lie on the same straight line.}
\displaystyle \therefore P,O\text{ and }Q\text{ are collinear.}
\displaystyle \therefore \text{The line joining the midpoints of two parallel chords passes through the center.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In the adjoining figure, }\widehat{AB}\cong\widehat{CD}.\text{ Prove that }\angle A=\angle B.
\displaystyle \text{Answer:} 2018-12-02_18-35-34\displaystyle \text{Given: }\widehat{AB}\cong\widehat{CD}.
\displaystyle \text{To prove: }\angle OAC=\angle OBD.
\displaystyle \text{Proof: Congruent arcs of a circle subtend equal angles at the center.}
\displaystyle \therefore \angle AOB=\angle COD
\displaystyle \text{Adding }\angle BOC\text{ to both sides,}
\displaystyle \angle AOB+\angle BOC=\angle COD+\angle BOC
\displaystyle \therefore \angle AOC=\angle BOD
\displaystyle \text{Consider }\triangle AOC\text{ and }\triangle BOD.
\displaystyle AO=OB\qquad\text{[Radii of the same circle]}
\displaystyle OC=OD\qquad\text{[Radii of the same circle]}
\displaystyle \angle AOC=\angle BOD\qquad\text{[Proved above]}
\displaystyle \therefore \triangle AOC\cong\triangle BOD\qquad\text{[By S.A.S. criterion]}
\displaystyle \therefore \angle OAC=\angle OBD\qquad\text{[By C.P.C.T.C.]}
\displaystyle \therefore \angle A=\angle B.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If two chords of a circle are equally inclined to the diameter through their}
\displaystyle \text{point of intersection, prove that the chords are equal.}
\displaystyle \text{Answer:} 2018-12-02_18-34-22\displaystyle \text{Given: Chords }AB\text{ and }AC\text{ are equally inclined to diameter }AD.
\displaystyle \therefore \angle OAL=\angle OAM
\displaystyle \text{To prove: }AB=AC.
\displaystyle \text{Construction: Draw }OL\perp AB\text{ and }OM\perp AC.
\displaystyle \text{Proof: Consider }\triangle AOL\text{ and }\triangle AOM.
\displaystyle \angle OLA=\angle OMA=90^\circ
\displaystyle \angle OAL=\angle OAM\qquad\text{[Given]}
\displaystyle AO=AO\qquad\text{[Common]}
\displaystyle \therefore \triangle AOL\cong\triangle AOM\qquad\text{[By A.A.S. criterion]}
\displaystyle \therefore OL=OM\qquad\text{[By C.P.C.T.C.]}
\displaystyle \text{Thus, chords }AB\text{ and }AC\text{ are equidistant from the centre }O.
\displaystyle \text{Chords of a circle equidistant from the centre are equal.}
\displaystyle \therefore AB=AC.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the adjoining figure, }O\text{ is the center of a circle and }PO\text{ bisects}
\displaystyle \angle APD.\text{ Prove that }AB=CD.
\displaystyle \text{Answer:} 2018-12-02_18-34-33\displaystyle \text{Given: }O\text{ is the center of the circle and }PO\text{ bisects }\angle APD.
\displaystyle \text{To prove: }AB=CD.
\displaystyle \text{Construction: Draw }OE\perp AB\text{ and }OF\perp CD,
\displaystyle \text{where }E\text{ and }F\text{ lie on the sides }PA\text{ and }PD,\text{ respectively.}
\displaystyle \text{Proof: Consider }\triangle OEP\text{ and }\triangle OFP.
\displaystyle \angle OEP=\angle OFP=90^\circ
\displaystyle \angle OPE=\angle OPF\qquad\text{[Since }OP\text{ bisects }\angle APD\text{]}
\displaystyle OP=OP\qquad\text{[Common]}
\displaystyle \therefore \triangle OEP\cong\triangle OFP\qquad\text{[By A.A.S. criterion]}
\displaystyle \therefore OE=OF\qquad\text{[By C.P.C.T.C.]}
\displaystyle \text{Thus, chords }AB\text{ and }CD\text{ are equidistant from the center }O.
\displaystyle \text{Chords of a circle equidistant from the center are equal.}
\displaystyle \therefore AB=CD.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Two equal chords }AB\text{ and }CD\text{ of a circle with center }O,\text{ when}
\displaystyle \text{produced, meet at a point }E\text{ as shown in the adjoining diagram. Prove that}
\displaystyle BE=DE\text{ and }AE=CE.
\displaystyle \text{Answer:} 2018-12-02_18-34-00\displaystyle \text{Given: }AB=CD.
\displaystyle \text{To prove: }BE=DE\text{ and }AE=CE.
\displaystyle \text{Construction: Draw }OL\perp AB\text{ and }OM\perp CD.
\displaystyle \text{Proof: Since equal chords of a circle are equidistant from the center,}
\displaystyle OL=OM.
\displaystyle \text{Consider }\triangle OLE\text{ and }\triangle OME.
\displaystyle \angle OLE=\angle OME=90^\circ
\displaystyle OE=OE\qquad\text{[Common hypotenuse]}
\displaystyle OL=OM\qquad\text{[Equal chords are equidistant from the center]}
\displaystyle \therefore \triangle OLE\cong\triangle OME\qquad\text{[By R.H.S. criterion]}
\displaystyle \therefore LE=ME\qquad\text{[By C.P.C.T.C.]}\qquad\ldots\text{(i)}
\displaystyle \text{The perpendicular from the center of a circle to a chord bisects the chord.}
\displaystyle \therefore BL=\frac{1}{2}AB\text{ and }DM=\frac{1}{2}CD
\displaystyle \text{But }AB=CD.
\displaystyle \therefore BL=DM\qquad\ldots\text{(ii)}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle LE-BL=ME-DM
\displaystyle \therefore BE=DE\qquad\ldots\text{(iii)}
\displaystyle \text{Now, }AE=AB+BE\text{ and }CE=CD+DE.
\displaystyle \text{Since }AB=CD\text{ and }BE=DE,
\displaystyle AB+BE=CD+DE
\displaystyle \therefore AE=CE.
\displaystyle \therefore BE=DE\text{ and }AE=CE.\text{ Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Prove that the line joining the midpoints of two equal chords of a circle}
\displaystyle \text{subtends equal angles with the chords.}
\displaystyle \text{Answer:} 2018-12-03_8-02-55\displaystyle \text{Given: }AB=CD,\text{ and }L\text{ and }M\text{ are the midpoints of }AB\text{ and }CD,
\displaystyle \text{respectively.}
\displaystyle \text{To prove: }\angle ALM=\angle CML\text{ and }\angle BLM=\angle DML.
\displaystyle \text{Construction: Join }OL\text{ and }OM.
\displaystyle \text{Proof: Since }L\text{ and }M\text{ are the midpoints of chords }AB\text{ and }CD,
\displaystyle OL\perp AB\text{ and }OM\perp CD.
\displaystyle \text{Since equal chords of a circle are equidistant from the center,}
\displaystyle OL=OM.
\displaystyle \therefore \triangle OLM\text{ is an isosceles triangle.}
\displaystyle \therefore \angle OLM=\angle OML\qquad\ldots\text{(i)}
\displaystyle \text{Now, }\angle BLM=90^\circ+\angle OLM
\displaystyle \text{and }\angle DML=90^\circ+\angle OML.
\displaystyle \text{Using (i),}
\displaystyle \angle BLM=\angle DML.
\displaystyle \text{Also, }\angle ALM=90^\circ-\angle OLM
\displaystyle \text{and }\angle CML=90^\circ-\angle OML.
\displaystyle \text{Using (i),}
\displaystyle \angle ALM=\angle CML.
\displaystyle \therefore \angle ALM=\angle CML\text{ and }\angle BLM=\angle DML.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the adjoining figure, }L\text{ and }M\text{ are the midpoints of two equal}
\displaystyle \text{chords }AB\text{ and }CD\text{ of a circle with center }O.\text{ Prove that:}
\displaystyle \text{(i) }\angle OLM=\angle OML\qquad\text{(ii) }\angle ALM=\angle CML.
\displaystyle \text{Answer:} 2018-12-03_8-10-28\displaystyle \text{Given: }AB=CD,\text{ and }L\text{ and }M\text{ are the midpoints of }AB\text{ and }CD,
\displaystyle \text{respectively.}
\displaystyle \text{To prove: (i) }\angle OLM=\angle OML\qquad\text{(ii) }\angle ALM=\angle CML.
\displaystyle \text{Construction: Join }OL\text{ and }OM.
\displaystyle \text{Proof: Since }L\text{ and }M\text{ are the midpoints of chords }AB\text{ and }CD,
\displaystyle OL\perp AB\text{ and }OM\perp CD.
\displaystyle \text{Since equal chords of a circle are equidistant from the center,}
\displaystyle OL=OM.
\displaystyle \text{Therefore, }\triangle OLM\text{ is an isosceles triangle.}
\displaystyle \therefore \angle OLM=\angle OML
\displaystyle \text{[Angles opposite equal sides of a triangle are equal]}
\displaystyle \therefore \text{(i) }\angle OLM=\angle OML.
\displaystyle \text{Now, since }OL\perp AB,
\displaystyle \angle ALM=90^\circ-\angle OLM.
\displaystyle \text{Similarly, since }OM\perp CD,
\displaystyle \angle CML=90^\circ-\angle OML.
\displaystyle \text{But }\angle OLM=\angle OML.
\displaystyle \therefore 90^\circ-\angle OLM=90^\circ-\angle OML
\displaystyle \therefore \text{(ii) }\angle ALM=\angle CML.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{PQ and }RQ\text{ are chords of a circle equidistant from the center. Prove}
\displaystyle \text{that the diameter passing through }Q\text{ bisects }\angle PQR\text{ and }\angle PSR.
\displaystyle \text{Answer:} 2018-12-02_18-33-15\displaystyle \text{Given: Chords }PQ\text{ and }RQ\text{ are equidistant from the center }O.
\displaystyle \text{Also, }QS\text{ is a diameter of the circle.}
\displaystyle \text{To prove: }\angle PQS=\angle RQS\text{ and }\angle PSQ=\angle RSQ.
\displaystyle \text{Construction: Join }PS\text{ and }RS.
\displaystyle \text{Proof: Chords of a circle equidistant from the center are equal.}
\displaystyle \therefore PQ=RQ
\displaystyle \text{Consider }\triangle PQS\text{ and }\triangle RQS.
\displaystyle \angle QPS=\angle QRS=90^\circ
\displaystyle \text{[Angles in semicircles with diameter }QS\text{]}
\displaystyle QS=QS\qquad\text{[Common hypotenuse]}
\displaystyle PQ=RQ\qquad\text{[Proved above]}
\displaystyle \therefore \triangle PQS\cong\triangle RQS\qquad\text{[By R.H.S. criterion]}
\displaystyle \therefore \angle PQS=\angle RQS\qquad\text{[By C.P.C.T.C.]}
\displaystyle \therefore QS\text{ bisects }\angle PQR.
\displaystyle \text{Also, }\angle PSQ=\angle RSQ\qquad\text{[By C.P.C.T.C.]}
\displaystyle \therefore QS\text{ bisects }\angle PSR.
\displaystyle \therefore \text{The diameter }QS\text{ bisects }\angle PQR\text{ and }\angle PSR.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If two chords of a circle bisect each other, show that they must be}
\displaystyle \text{diameters.}
\displaystyle \text{Answer:} 2018-12-02_18-32-51\displaystyle \text{Given: Chords }AB\text{ and }CD\text{ intersect at }O\text{ and bisect each other.}
\displaystyle \therefore OA=OB\text{ and }OC=OD.
\displaystyle \text{To prove: }AB\text{ and }CD\text{ are diameters of the circle.}
\displaystyle \text{Construction: Join }AC,CB,BD\text{ and }DA.
\displaystyle \text{Proof: Consider }\triangle AOC\text{ and }\triangle BOD.
\displaystyle OA=OB\qquad\text{[Given]}
\displaystyle OC=OD\qquad\text{[Given]}
\displaystyle \angle AOC=\angle BOD\qquad\text{[Vertically opposite angles]}
\displaystyle \therefore \triangle AOC\cong\triangle BOD\qquad\text{[By S.A.S. criterion]}
\displaystyle \therefore AC=BD\qquad\text{[By C.P.C.T.C.]}
\displaystyle \therefore \widehat{AC}\cong\widehat{BD}\qquad\ldots\text{(i)}
\displaystyle \text{[Equal chords of a circle cut off equal arcs]}
\displaystyle \text{Now, consider }\triangle AOD\text{ and }\triangle BOC.
\displaystyle OA=OB\qquad\text{[Given]}
\displaystyle OD=OC\qquad\text{[Given]}
\displaystyle \angle AOD=\angle BOC\qquad\text{[Vertically opposite angles]}
\displaystyle \therefore \triangle AOD\cong\triangle BOC\qquad\text{[By S.A.S. criterion]}
\displaystyle \therefore AD=BC\qquad\text{[By C.P.C.T.C.]}
\displaystyle \therefore \widehat{AD}\cong\widehat{BC}\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle \widehat{AC}+\widehat{AD}\cong\widehat{BD}+\widehat{BC}
\displaystyle \therefore \widehat{CAD}\cong\widehat{CBD}
\displaystyle \text{These two arcs together form the complete circle. Therefore, each is a semicircle.}
\displaystyle \therefore CD\text{ is a diameter.}
\displaystyle \text{Similarly,}
\displaystyle \widehat{AC}+\widehat{BC}\cong\widehat{BD}+\widehat{AD}
\displaystyle \therefore \widehat{ACB}\cong\widehat{BDA}
\displaystyle \text{Therefore, each of these arcs is a semicircle.}
\displaystyle \therefore AB\text{ is a diameter.}
\displaystyle \therefore AB\text{ and }CD\text{ are diameters of the circle. Hence proved.}
\displaystyle \\

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