Question 21: Prove that: (i) the parallelogram inscribed in a circle is a rectangle (ii) the rhombus, inscribed in a circle is a square.

(i) Let $ABCD$ be a parallelogram inscribed in the circle.

$\therefore \angle BAD = \angle BCD$ (opposite angles of a parallelogram are equal)

$\angle BAD + \angle BCD = 180^o$

$\Rightarrow \angle BAD = \angle BCD = 90^o$

Similarly, $\angle ABC = \angle ADC = 90^o$

Therefore $ABCD$ is a rectangle

(ii) $ABCD$ is a rhombus (given) i.e. all four sides are equal.

$\angle BAD = \angle BCD = 90^o$

Similarly, $\angle ABC = \angle ADC = 90^o$

Therefore $ABCD$ is a square

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Question 22: In the following figure $AB = AC$. Prove that $DEBC$ is an isosceles trapezium.

$AB = AC$ (given)

$\angle ABC = \angle ACB$

$\angle CBA+\angle CED = 180^o (BCED$ is a cyclic quadrialteral)

$\angle ACB + \angle CED = 180^o$

$DE \parallel BC$ … … … … (i)

$\angle ADE = \angle ABC$ (corresponding angles)

$\angle AED = \angle ACB$

Hence $\angle ADE = \angle AED$

$\Rightarrow AD = AE$

$AB-AD = AC-AE$

$\Rightarrow BD = EC$ … … … … (ii)

Hence because of (i) and (ii), $ABCD$ is an isosceles trapezium.

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Question 23: Two circles intersect at $P$ and $Q$. Through $P$ diameters $PA$ and $PB$ of the two circles are drawn. Show that the points $A, Q$ and $B$ are collinear.

$\angle PQA = 90^o$ (angle in a semi circle)

$\angle PQB = 90^o$ (angle in a semi circle)

Therefore $\angle AQB = 180^o$

Therefore $A, Q \ and \ B$ are collinear.

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Question 24: $ABCD$ is a quadrilateral inscribed in a circle. having $\angle A = 60^o$. $O$ is the center of the circle. Show that: $\angle OBD+\angle ODB= \angle CBD+\angle CDB$.

$\angle BAD + \angle BCD = 180^o$ (cyclic quadrilateral)

$\angle BCD = 180^o-60^o=120^o$

$\angle BOD = 2 \angle BAD$

$\angle CBD + \angle CDB = 180^o-120^o = 60^o$ (sum of the angles in a triangle is 180)

Similarly $\angle OBD + \angle ODB = 180^o-120^o=60^o$

Therefore $\angle OBD + \angle ODB = \angle CBD + \angle CDB$

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Question 25: The figure given below shows the circle with center $O$. Given $\angle AOC = a$ and $\angle ABC=b$. (i) Find the relationship between $a \ and \ b$. (ii) Find $\angle OAB$ if $OABC$ is a parallelogram.

(i) $\angle AOC = a$ and $\angle CBA = b$ (given)

$\angle ABC = \frac{1}{2} Reflex (\angle COA)$

$b = \frac{1}{2} (360^o-a)$

$2b = 360^o-a \ or \ a+2b=360^o$

(ii) If $OABC$ is a parallelogram

Therefore $a = b$ (opposite angles are equal)

$\therefore 3b = 360^o \Rightarrow b = 120^o$

$\therefore a = 120^o$

$\angle ACB = \angle BAC$

$\therefore \angle BAC = 30^o$

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

$\therefore \angle OCA = \angle OAC$

$\therefore \angle OAC = 30^o$

Hence $\angle OAB = 30^o+30^o=60^o$

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Question 26: Two chords $AB$ and $CD$ intersect at $P$ inside the circle. Prove that the sum of the angles subtended by the arcs $AC$ and $BD$ at the center $O$ is equal to twice the $\angle APC$.

$\angle AOC = 2 \angle ADC$

Similarly, $\angle BOD = 2 \angle BAD$

$\angle AOC + \angle BOD = 2(\angle ADC+\angle BAD)$ … … … … (i)

In  $\triangle PAD$

$180^o-\angle APC + \angle PAD + \angle ADC = 180^o$

$\therefore \angle PAD + \angle ADC = \angle APC$ … … … … (ii)

Using (1) and (ii)

$\angle AOC + \angle BOD = 2(\angle BAD + \angle ADC) = 2 \angle APC$

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Question 27: In the given figure $RS$ is a diameter of the circle. $NM \parallel RS$ and $\angle MRS= 29^o$. Find (i) $\angle RNM$ (ii) $\angle NRM$

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

$\therefore \angle RMS = 180^o-90^o-29^o = 61^o$

$\therefore \angle RNM + 61^o = 180^o \Rightarrow \angle RNM = 119^o$

(ii) Given $NM \parallel RS$

$\angle NMR = \angle MRS = 29^o$ (alternate angles)

$\angle NMS = 90^o+29^o = 119^o$

$\angle NRS + \angle NMS = 180^o$

$\angle NRM+\angle MRS + \angle NMR + \angle RMS = 180^o$

$\angle NRM +29^o+29^o+90^o = 180^o$

$\Rightarrow \angle NRM = 180^o-148^o = 32^o$

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Question 28: In the given figure, $AB \parallel CD$ and $O$ is the center of the circle. If $\angle ADC = 25^o$, find the $\angle AEB$. Give reasons.

$\angle CAD = 90^o$ and $\angle CBD = 90^o$ (angles in a semi circle)

Since $AB \parallel CD$

$\angle CDA = \angle DAB = 25^o$

$\angle BAC = \angle BAD + \angle CAD$

$= 25^o+90^o=115^o$

$\therefore \angle ADB = 180^o-25^o -\angle ABD$

In $\triangle \angle ACD +90^o+25^o = 180^o$

$\Rightarrow \angle ACD = 180^o-115^o = 65^o$

$\angle ACD + \angle ABD = 180^o$ (cyclic quadrilateral)

$\angle ABD = 180^o-65^o=115^o$

$\angle ADB = 180^o-25^o-115^o=40^o$

$\therefore \angle ADB = \angle AEB = 40^o$ (angles in the same segment)

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Question 29: Two circles intersect at $P$ and $Q$. Through $P$ a straight line $APB$ is drawn to meet the circles in $A$ and $B$. Through $Q$, a straight line is drawn to meet the circles at $C$ and $D$. Prove that $AC$ is parallel to $BD$.

$APQC$ and $BPQD$ are cyclic quadrilateral

$\angle CAP + \angle PQC = 180^o$ … … … … (i)

$\angle PQD + \angle PBD = 180^o$ … … … … (ii)

$\angle PQC + \angle PQD = 180^o$  … … … … (iii)

From (i) and (ii)

$\angle CAP + 180^o - \angle PQD = 180^o$

$\angle CAP = \angle PQD$  … … … … (iv)

From (iv) and (iii)

$\angle CAP = 180^o -\angle PBD$

$\Rightarrow \angle CAP + \angle PBD = 180^o$    … … … … (v)

Therefore $AB \parallel BD$ by (v)

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Question 30: $ABCD$ is a cyclic quadrilateral in which $AB$ and $DC$ on being produced, meet at $P$ such that $PA = PD$. Prove that $AD$ is parallel to $BC$.

$ABCD$ is a cyclic quadrilateral, $AD \parallel BC$ and $PA = PD$ (given)

$\angle PAD = \angle PDA$

$\angle CDA = 180^o - \angle PDA$

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

$\angle ABC + 180^o - \angle PAD = 180$

$\angle ABC = \angle PAD$

$\therefore AD \parallel BC$

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