Question 1: In the given circle with diameter $AB$, find the value of $x$. [2003]

$\angle ABD = \angle ACD = 30^o$ (angles in the same segment)

In $\triangle ADB$

$\angle BAD + \angle ADB + \angle ABD = 180^o$

Since $AB$ is the diameter,

$\angle ADB = 90^o$ (angle in the semi circle)

$x+ 90+30 = 180 \Rightarrow x = 60^o$

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Question 2: In the given figure, $O$ is the center of the circle with radius $5\ cm$. $OP$ and $OQ$ are $\perp$ to $AB$ and $CD$ respectively. $AB = 8\ cm$ and $CD = 6\ cm$. Determine the length of $PQ$.

Given: $Radius = 5 \ cm, \ OP \perp AB \ and \ OQ \perp CD$

$AB = 8 \ cm \ and \ CD = 6 \ cm$

$P \ and \ Q$ are mid point of $AB \ and \ CD$ respectively ($\perp$ from the center to a chord will bisect the chord)

In $\triangle OAP$

$OA^2= OP^2+AP^2$

$25 = OP^2 + 16 \Rightarrow OP = 3 \ cm$

Similarly, in $\triangle OCQ$

$OC^2= OQ^2+CQ^2$

$25 = OQ^2+9 \Rightarrow OQ = 4 \ cm$

Hence $PQ = OP + OQ = 3 + 4 = 7 \ cm$

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Question 3: The given figure shows two circles with centers $A$ and $B$; and radius $5\ cm$ and $3\ cm$ respectively, touching each other internally. If the perpendicular bisector of $AB$ meets the bigger circle in $P$ and $Q$, find the length of $PQ$.

$AB = AC - BC = 5 - 3 = 2 \ cm$

$PQ$ bisects $AB$ (let the point be $R$)

$AR = 1 \ cm$

In $\triangle APR$

$AP^2= RP^2+AR^2$

$\Rightarrow 25 = RP^2+1 \Rightarrow RP = 24 = 2\sqrt{6} \ cm$

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Question 4: In the given figure, $\triangle ABC$ in which $\angle BAC = 30^o$. Show that $BC$ is equal to the radius of the circumcircle of the $\triangle ABC$, whose center is $O$.

Given $\angle BAC = 30^o$

$\angle BOC = 2 \angle BAC = 2 . 30 = 60^o$ (angle at the center of the circle is twice that of the angle subtend at the circumference by the same chord)

In $\triangle OBC$

$OB = OC$  (radius of the same circle)

$\angle OBC + \angle OCB + \angle BOC = 180^o$

$\angle OBC + \angle OCB = 180-60 = 120^o$

$2 \angle OBC = 120^o \Rightarrow \angle OBC = 60^o$

Therefore $\angle OBC = \angle OCB = \angle BOC = 60^o$

Therefore $\triangle BOC$ is equilateral

$\Rightarrow BC = OB = OC$

Hence $BC$ is equal to the radius of the circle.

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Question 5: Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Given: Isosceles $\triangle ABC$, where $AB = AC$.

$AB$ is the diameter of the circle.

$\angle ADB = 90^o$ (angle in the semi circle)

$\angle ADC = 180-90=90^o$

Consider $\triangle ADB$ and $\triangle ADC$

$AD$ is common

$\angle ADC = \angle ADB$

$AB = AC$

Therefore  $\triangle ADB \cong \triangle ADC (SAS \ axiom)$

Therefore $BD = DC$

Hence $D$ is the midpoint of $BC$.

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Question 6: In the given figure, chord $ED$ is parallel to diameter $AC$ of the circle. Given $\angle CBE = 65^o$, calculate $\angle DEC$.

Given $AD \parallel AC$

$\angle CBE = 65^o$

$\widehat{EC}$ subtends $\angle EOC$ at center  and $\angle EBC$ at circumference.

$\angle ECO = 2 \angle EBC = 2 \times 65 = 130^o$

In $\triangle EOC$,

$EO = OC$ (radius of the same circle)

$\angle OEC = \angle OCE$

$\angle OEC + \angle OCE + \angle EOC = 180^o$

$2 \angle OEC + 130 = 180 \Rightarrow \angle OEC = 25^o$

Given $ED \parallel AC$

$\Rightarrow \angle DEC = \angle OEC \Rightarrow \angle DEC = 25^o$

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Question 7: Chords $AB$ and $CD$ of a circle intersect each other at point $P$ such that $AP = CP$. Show that: $AB = CD$.

Given $AB$ and $CD$ intersect at $P$

$AP = CP$ (given)

Therefore $AP \times PB = CP \times PD$

$\Rightarrow \frac{AP}{CP} = \frac{PD}{PB}$

$\Rightarrow PD = PB$

Therefore $AP + PB = CP + PD \Rightarrow AB = CD$

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Question 8: The quadrilateral formed by the angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

Given: $ABCD$ is a cyclic quadrilateral and $PQRS$ is a quadrilateral

In $\triangle APD$:

$PAD + ADP + APD = 180^o$ … … … … … (i)

In $\triangle BQC$

$QBC + BCQ + BQC = 180^o$  … … … … … (ii)

$PAD + ADP + APD + QBC + BCQ + BQC = 360^o$

$\frac{1}{2} (BAD + ADC + DCB + CBA) + APD + BQC = 360^o$

$180 + APD + BQC = 360^o \Rightarrow APD + BQC = 180^o$

Therefore $PQRS$ is a cyclic quadrilateral as the opposite angles are supplementary.

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Question 9: In the given diagram, $\angle DBC = 58^o$, $BD$ is a diameter of the circle. Calculate: (i) $\angle BDC$ (ii) $BEC$ (iii) $\angle BAC$ [2014]

Given $BD$ is diameter

$\angle DBC = 58^o$

(i) $\angle DCB = 90^o$ (angle in semi circle)

In $\triangle BDC$

$58+\angle BDC+90=180^o$

$\angle BDC = 32^o$

(ii) $ABEC$ is a cyclic quadrilateral

$\angle BDC = \angle BAC = 32^o$ (angle in the same segment)

Therefore $\angle BEC + 32 = 180^o$

$\angle BEC = 148^o$

(iii) $\angle BAC = 32^o$ (angle in the same segment)

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Question 10: $D$ and $E$ are points on equal sides $AB$ and $AC$ of an isosceles $\triangle ABC$ such that $AD = AE$. Prove that the points $B, C, E$ and $D$ are concyclic.

In $\triangle ABC AB = AC$. Also given $AD = AE$

Therefore $\angle ABC = \angle ACB = x \ (say)$

In $\triangle ADE, AD = AE$

Now in $\triangle ABC$

$\frac{AD}{AB} =\frac{AE}{AC}$

$\Rightarrow DE \parallel BC$

$\Rightarrow \angle BCE = \angle DEA = \angle ADE = x$

Therefore $\angle BCE + \angle BDE = x + (180-x) = 180$

$\Rightarrow BDCE$ is a cyclic quadrilateral.

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Question 11: In the given figure, $ABCD$ is a cyclic quadrilateral. $AF \parallel CB$ and $DA$ is produced to point $E$. If $\angle ADC = 92^o, \angle FAE= 20^o$; Determine $\angle BCD$. Give reason in support of your answer.

Given: $ABCD$ is a cyclic quadrilateral

$AF \parallel CB. \angle ADC = 92^o$ and $\angle FAE = 20^o$

$\angle CDA + \angle CBA = 180^o$

$\angle CBA = 180-92 = 88^o$

Since $AF \parallel CB$

$\angle CBA = \angle BAF$ (alternate angles)

$\angle BAE = 88 + 20 = 108^o$

Therefore $\angle DBA = 180-108 = 72^o$

Therefore $\angle BCD = 180-72 = 108^o$

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Question 12: If $I$ is the incentre $\triangle ABC$ and $AI$ when produced meets the circumcircle of $\triangle ABC$ at point $D$. If $\angle BAC = 66^o$ and $\angle ABC = 80^o$. Calculate (i) $\angle DBC$, (ii) $\angle IBC$, (iii) $\angle BIC$

Given $\angle BAC = 66^o$ and $\angle ABC = 80^o$

(i) $\angle DBC = \angle DAC$ (angles in the same segment)

Since $I$ is the incenter

$\angle DAC = \frac{1}{2} \angle BAC = \frac{1}{2} \times 66 = 33^o$

Therefore $\angle DBC = 33^o$

(ii) Similarly, $\angle IBC = \frac{1}{2} \angle ABC = \frac{1}{2} \times 80 = 40^o$

(iii) In $\triangle ABC, \angle ACB = 180-\angle ABC - \angle BAC = 180-80-66 = 34^o$

Also $\angle ICB = \frac{1}{2} \angle BCA = \frac{1}{2} \times 34 = 17$

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Question 13: ln the given figure, $AB = AD = DC = PB$ and $\angle DBC = x^o$. Determine, in terms of $x$ : (i) $\angle ABD$, (ii) $\angle APB$. Hence or otherwise, prove that $AP \parallel DB$.

Given $AB = AD =DC = PB$

$\angle DBC = x^o$

$\angle DAC = \angle DBC$ (angles in the same segment)

Since $AC = DC$

$\angle DAC = \angle DCA = x$

Similarly, $\angle ABD = \angle DAC$ ( $Since ABD = ACD \Rightarrow ACD = DAC$)

In $\triangle ABP, \ \angle ABC = \angle BAP + \angle APB$

Since $AB = BP$

$\angle BAP = \angle APB$

$2x = \angle APB + \angle APB = 2 \angle APB$

$\Rightarrow \angle APB = x^o$

Therefore $\angle APB = \angle DBC = x^o$ and hence $AP \parallel DB$ (alternate angles)

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Question 14: In the given figure; $ABC, AEQ$ and $CEP$ are straight lines. Show that $\angle APE$ and $\angle CQE$ are supplementary.

Given $ABC, AEQ \ and \ CEP$ are straight lines.

In cyclic quadrilateral $ABFP$

$\angle APE + \angle ABE = 180^o$ … … … … … (i)

Similarly in cyclic quadrilateral $BCQE$

$\angle CQE+\angle CBE = 180^o$ … … … … … (ii)

$\angle APE + \angle ABE + \angle CQE+\angle CBE = 360$

$\angle APE + \angle CQE + 180 = 360$

$\Rightarrow \angle APE + \angle CQE = 180$

Hence they are supplementary.

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