Question 15: In the given figure, $AB$ is the diameter of the circle with center $O$. If $\angle ADC = 32^o$, find $\angle BOC$.

$\widehat{AC}$ subtends $AOC$ at the center and $ADC$ at the circumference of the circle.

$\angle AOC = 2 \angle ADC$

$\Rightarrow \angle AOC = 2 \times 32 = 64^o$

$\angle AOC + \angle BOC = 180 \Rightarrow \angle BOC = 180-64=116^o$

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Question 16: In a cyclic-quadrilateral $PQRS, \angle PQR = 135^o$. Sides $SP$ and $RQ$ produced meet at point $A$ whereas sides $PQ$and $SR$ produced meet at point $B$. lf $\angle A: \angle B = 2 : 1$; find $\angle A$ and $\angle B$.

Given $PQRS$ is a cyclic quadrilateral.

$\angle A = \angle B = 2 :1$  or if $\angle B = x$, then $\angle A = 2x$

$\angle PQR +\angle PSR = 180^o$ (sum of the opposite angles in a cyclic quadrilateral is $180^o$ )

$\angle PSR = 180-135 = 45^o$

In $\triangle PBS; \angle SPB + 45+x = 180^o$

$\Rightarrow \angle SPB = 135-x$ … … … (i)

In $\triangle PQA; \angle APQ = 180-2x-45 = 135-2x$

Therefore $\angle SPB = 180 - (135-2x) = 45+2x$  … … … (ii)

Hence $35-x = 45+2x \Rightarrow x = 30^o$

Therefore $\angle B = 30^o \ and \ \angle A = 60^o$

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Question 17: In the following figure, $AB$ is the diameter of a circle with center $O$ and $CD$ is the chord with length equal to radius $OA$. If $AC$ produced and $BD$ produced meet at point $P$; show that $\angle APB = 60^o$.

Given: $AB$ is diameter, $CD = OA = Radius$

In $\triangle CDO, CD = OD = OC$

$\Rightarrow \triangle CDO$ is equilateral

Therefore $\angle OCD = \angle OCD = \angle COD = 60^o$

In $\triangle AOC$

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

Therefore $\angle OCA = \angle CAO$

Similarly in $\triangle BOD$

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

Therefore $\angle ODB = \angle BDO$

Since $ABCD$ is cyclic quadrilateral

Therefore $\angle ACD + \angle OBD = 180^o$ (opposite angles of a cyclic quadrilateral are supplementary)

$60+ \angle ACO + \angle OBD = 180^o$

$\Rightarrow \angle ACO + \angle OBD = 120^o$

In $\triangle APB$

$\angle APB + \angle PBA + \angle BAP = 180^o$

$\angle APB + 120 = 180^o$

$\Rightarrow \angle APB = 60^o$. Hence proved.

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Question 18: In the following figure, $ABCD$ is a cyclic quadrilateral in which $AD$ is parallel to $BC$. If the bisector of $\angle A$ meets $BC$ at point $E$ and the given circle at point $F$, prove that: (i) $EF = FC$ (ii) $BF = DF$

Given: $ABCD$ is a cyclic quadrilateral

$AD \parallel BC$

$\angle BAE = \angle EAD \ (AE$ is the angle bisector) … … … (i)

(i) $\angle AEB = \angle EAD$ (alternate angles) … … … (ii)

In $\triangle ABE$

$\angle ABE = 180^o - \angle BAE - \angle AEB$

Using (i) and (ii) we get  $\angle ABE = 180^o - 2 \angle AEB$

$\angle CEF = \angle AEB$ (vertically opposite angles)

$\angle ADC = 180^o - \angle ABC = 180^o -(180^o-2 \angle AEB)$

$\Rightarrow \angle ADC = 2 \angle AEB$

$\angle AFC = 180^o - \angle ADC = 180^o - 2 \angle AEB (since AFCD$ is a cyclic quadrilateral)

$\angle ECF = 180^o - (\angle AFC + \angle CEF) \\ = 180^o - (180^o-2 \angle AEB + \angle AEB) = \angle AEB$

Also $\angle AEB = \angle CEF \Rightarrow EF = EC$

(ii) $\widehat{BF}$ subtends $\angle BAF$ on circumference

$\widehat{OF}$ subtends $\angle FAD$ on circumference

Given $\angle BAF = \angle FAD$

Therefore $\widehat{BF} = \widehat{OF}$ (equal arcs subtends equal angles on circumference)

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Question 19: $ABCD$ is a cyclic quadrilateral. Sides $AB$ and $DC$ produced meet at point $E$; whereas sides $BC$ and $AD$produced meet at point $F$. If $\angle DCF: \angle F: \angle E = 3: 5 : 4$, find the angles of the cyclic quadrilateral $ABCD$.

Given $DCF : F : E = 3:4:5$

In $\triangle CDF: \angle ADC = 3x+5x = 8x$

In $\triangle BCE: \angle ABC = 3x+4x = 7x$

$\angle ABC + \angle ADC = 180^o$

or $7x+8x= 180 \Rightarrow x = 12^o$

Therefore

$\angle DAC = 180 -12x = 180 - 144 = 36^o$

$\angle CBA = 7x = 7 \times 12 = 84^o$

$\angle DCB = 180-3x = 180-36 = 144^o$

$\angle ADC = 8x = 8 \times 12 = 96^o$

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Question 20: In the following figure shows a circle with $PR$ as its diameter. If $PQ =7\ cm$ and $QR=3RS=6cm$, find the perimeter of the cyclic quadrilateral $PQRS$. [1992]

Given: $PQ = 7 \ cm, QR = 3RS = 6 \ cm$

$PR = \sqrt{7^2+6^2} = \sqrt{85}$

$SR = \sqrt{PR^2-PS^2} = \sqrt{85-4} =9$

Therefore the perimeter of $PQRS = 2+9+6+7 = 24 \ cm$

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Question 21: In the following figure, $AB$ is the diameter of a circle with center $O$. If $chord AC = chord AD$, prove that: (i) $arc BC = arc DB$ (ii) $AB$ is bisector of $\angle CAD$. Further, if the length of $arc AC = 2 \times arc BC$, find : (a) $\angle BAC$ (b) $\angle ABC$.

Given: $AC = AD$

Consider $\triangle ABC$ and $\triangle ABD$

$AB$ is common

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

$AC = AD$

Therefore $\triangle ABC \cong \triangle ABD$

(i) $BC = BD$ (corresponding parts of congruent triangles)

(ii) $\angle CAB = \angle BAD$ (equal chords subtend equal angles on the circumference of the same circle)

Therefore $AB$ bisects $\angle CAD$

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Question 22: In cyclic quadrilateral $ABCD$; $AD = BC, \angle BAC = 30^o$ and $\angle CBD = 70^o$; find: (i) $\angle BCD$ (ii) $\angle BCA$ (iii) $\angle ABC$ (iv) $\angle ADC$

Given $AD = BC, \angle BAC = 30^o$ and $\angle CBD = 70^o$

$\angle BDC = \angle BAC = 30^o$ (angles in the same segment)

$\angle DAC = \angle DBC = 70^o$ (angles in the same segment)

$\angle BCD + 30^o + 70^o = 180^o \Rightarrow \angle BCD = 80^o$ (opposite angles in a cyclic quadrilateral are supplementary)

$BC = AD \Rightarrow \angle BDC = \angle ADC = 30^o$

Therefore $\angle BCA = 50^o$

$70^o + 30^o + 30^o +\angle BDA = 180 \Rightarrow \angle BDA = 50^o$

$50^o+ 30^o + 70^o+\angle ABD = 180 \Rightarrow \angle ABD = 30^o$

$\angle ABC = 100^o, \angle BAD = 100^o, \angle BCD = 80^o \ and \ \angle CDA = 80^o$

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Question 23: In the given figure $\angle ACE = 43^o$ and $\angle CAF = 62^o$; find the values of $a, b$ and $c$. [2007]

Given $\angle ACE = 43^o$ and $\angle CAF = 62^o$

$\angle BAE + \angle BDE = 180^o$ (ABDE is a cyclic quadrilateral)

$\angle BDE = 118^o$

$\angle c + 118^o = 180^o$ (straight line) $\Rightarrow c = 62^o$

In $\triangle ACE$:

$62^o+43^o+\angle CEA = 180^o \Rightarrow \angle CEA = 75^o$

Therefore $\angle DEF = 180^o - 75^o = 105^o$

Therefore $\angle a + 75^o = 180^o \Rightarrow \angle a = 105^o \ (ABDE$ is a cyclic quadrilateral)

In $\triangle DEF$:

$62^o+ 105^o + \angle b = 180^o \Rightarrow \angle b = 13^o$

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Question 24: In the given figure, $AB \parallel DC, \angle BCE = 80^o$ and $\angle BAC = 25^o$. Find (i) $\angle CAD$ (ii) $\angle CBD$ (iii) $\angle ADC$

Given $\angle = 80^o$ and $\angle BAC = 25^o$. Also $AB \parallel DC$

Therefore $\angle CAB = \angle ACD$ (alternate angles)

Hence $\angle ACD = 25^o$

Therefore $\angle ACB = 180^o - 80^o - 25^o = 75^o$

and $\angle DCB = 25^o + 75^o = 100^o$

$\angle DAB = 180^o - 100^o = 80^o$

Therefore $\angle DAC = 80^o - 25^o = 55^o$

In $\triangle ABC$:

$\angle ABC = 180^o - 25^o - 75^o = 80^o$

Therefore $\angle ADC = 180^o = 80^o = 100^o$

$\angle CBD = \angle DAC = 55^o$ (angles in the same segment)

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Question 25: $ABCD$ is a cyclic quadrilateral of a circle center $O$ such that $AB$ is a diameter of the circle and the length of the chord $CD$ is equal to the radius of the circle. If $AD$ and $BC$ produced meet at $P$, show that $\angle APB = 60^o$.

Given $CD = OC = Radius$

In $\triangle OCD$:

$OD = OC = CD \Rightarrow \angle ODC \\ = \angle OCD = \angle DOC = 60^o$

In $\triangle AOD, OA = OD$ (Radius of the same circle)

Let $\angle OAD = \angle ADO = x$

Therefore $\angle AOD = 180^o -2x$ … … … (i)

In $\triangle COB, OC = OB$ (Radius of the same circle)

Let $\angle OBC = \angle OCB = y$

Therefore $\angle COB = 180^o - 2y$ … … … (ii)

Now, $180^o - 2x + 60^o + 180^o - 2y = 180^o$ (straight line angle)

$\Rightarrow x + y = 120^o$

In $\triangle APB$:

$x + y + \angle APB = 180^o \Rightarrow \angle APB = 180^o - 120^o = 60^o$

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Question 26: In the figure, given alongside, $CP$ bisects $\angle ACB$. Show that $DP$ bisects $\angle ADB$

Given $CP$ bisects $\angle ACB$

Therefore $\angle ACP = \angle PCB$

$\angle ACP = \angle ADP$ (angles in the same segment)

$\angle PCB = \angle PDB$ (angles in the same segment)

Therefore $\angle ADP = \angle PDB$

Hence $DP \ bisects \ \angle ADB$

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Question 27: In the figure shown, $AD = BC, \angle BAC = 30^o$ and $\angle CBD=70^o$. Find (i) $\angle BCD$ (ii) $\angle BCA$ (iii) $\angle ABC$ (iv) $\angle ADB$

From $chord \ CD: \angle DBC = \angle DAC$ ( angles in the same segment)

$\Rightarrow \angle DAC = 70^o$

From $chord \ CB: \angle CAB = \angle CDB$ (angles in the same segment)

$\Rightarrow \angle CDB = 30^o$

In $\triangle \angle DCB: 70 + 30 + \angle DCB = 180 \Rightarrow \angle DCB = 80^o$

Since $AD = BC: \angle DCA = \angle CAB \Rightarrow \angle DCA = 30^o$

In $\triangle ADC: 70 + \angle ADB + 30 + 30 = 180 \Rightarrow \angle ADB = 50^o$

$ABCD$ is a cyclic quadrilateral. $\therefore 50 + 30 + \angle DBA + 70 = 180 \Rightarrow DBA = 30^o$

Hence

(i) $\angle BCD = 80^o$

(ii) $\angle BCA = 50^o$

(iii) $\angle ABC = 100^o$

(iv) $\angle ADB = 50^o$

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Question 28: In the figure, given below, $AB$ and $CD$ are two parallel chords and $O$ is the center. If the radius of the circle is $15\ cm$, find the distance $MN$ between the two chords of lengths $24\ cm$ and $18\ cm$ respectively. [2010]

We know that perpendicular drawn from the center of the circle will bisect the chord.

Therefore $MO^2 = 15^2-12^2 \\ = 225 - 144 = 81 \Rightarrow MO = 81 \ cm$

$ON^2 = 15^2-9^2 = 225 - 81 = 12 \\ \Rightarrow ON = 12 \ cm$

Hence $MN = 9 + 12 = 21 \ cm$

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