Exercise 9.3 Practice
Angles Subtended by Arcs and Cyclic Quadrilaterals
Q1: Angles at Centre
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In the figure below, $A, B$ and $C$ are three points on a circle with centre $O$ such that $\angle BOC = 30^\circ$ and $\angle AOB = 60^\circ$. If $D$ is a point on the circle other than the arc $ABC$, find $\angle ADC$.
$\angle AOC = \angle AOB + \angle BOC$
$\angle AOC = 60^\circ + 30^\circ = 90^\circ$.
$\angle AOC = 60^\circ + 30^\circ = 90^\circ$.
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$\therefore \angle AOC = 2 \angle ADC$.
$\therefore \angle AOC = 2 \angle ADC$.
$\angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} (90^\circ) = 45^\circ$.
$\angle ADC = 45^\circ$
Q2: Chord Equal to Radius
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A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Since Chord $AB = \text{Radius } OA = \text{Radius } OB$, $\triangle OAB$ is an equilateral triangle.
$\therefore \angle AOB = 60^\circ$.
$\therefore \angle AOB = 60^\circ$.
Angle on Major Arc ($\angle APB$):
$\angle APB = \frac{1}{2} \angle AOB = \frac{1}{2}(60^\circ) = 30^\circ$.
$\angle APB = \frac{1}{2} \angle AOB = \frac{1}{2}(60^\circ) = 30^\circ$.
Angle on Minor Arc ($\angle AQB$):
$APBQ$ is a cyclic quadrilateral.
$\angle APB + \angle AQB = 180^\circ$.
$30^\circ + \angle AQB = 180^\circ \Rightarrow \angle AQB = 150^\circ$.
$APBQ$ is a cyclic quadrilateral.
$\angle APB + \angle AQB = 180^\circ$.
$30^\circ + \angle AQB = 180^\circ \Rightarrow \angle AQB = 150^\circ$.
Major Arc: $30^\circ$, Minor Arc: $150^\circ$
Q3: Reflex Angle
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In the figure below, $\angle PQR = 100^\circ$, where $P, Q$ and $R$ are points on a circle with centre $O$. Find $\angle OPR$.
The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Reflex $\angle POR = 2 \angle PQR = 2 \times 100^\circ = 200^\circ$.
Reflex $\angle POR = 2 \angle PQR = 2 \times 100^\circ = 200^\circ$.
Angle inside $\triangle POR$:
$\angle POR = 360^\circ - 200^\circ = 160^\circ$.
$\angle POR = 360^\circ - 200^\circ = 160^\circ$.
In $\triangle OPR$, $OP = OR$ (Radii).
$\therefore \angle OPR = \angle ORP$.
$\angle OPR + \angle ORP + \angle POR = 180^\circ$.
$2\angle OPR + 160^\circ = 180^\circ$.
$2\angle OPR = 20^\circ \Rightarrow \angle OPR = 10^\circ$.
$\therefore \angle OPR = \angle ORP$.
$\angle OPR + \angle ORP + \angle POR = 180^\circ$.
$2\angle OPR + 160^\circ = 180^\circ$.
$2\angle OPR = 20^\circ \Rightarrow \angle OPR = 10^\circ$.
$\angle OPR = 10^\circ$
Q4: Angles in Same Segment
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In the figure below, $\angle ABC = 69^\circ$, $\angle ACB = 31^\circ$, find $\angle BDC$.
In $\triangle ABC$:
$\angle BAC + \angle ABC + \angle ACB = 180^\circ$.
$\angle BAC + 69^\circ + 31^\circ = 180^\circ$.
$\angle BAC + 100^\circ = 180^\circ \Rightarrow \angle BAC = 80^\circ$.
$\angle BAC + \angle ABC + \angle ACB = 180^\circ$.
$\angle BAC + 69^\circ + 31^\circ = 180^\circ$.
$\angle BAC + 100^\circ = 180^\circ \Rightarrow \angle BAC = 80^\circ$.
Angles in the same segment of a circle are equal.
$\therefore \angle BDC = \angle BAC$.
$\angle BDC = 80^\circ$.
$\therefore \angle BDC = \angle BAC$.
$\angle BDC = 80^\circ$.
$\angle BDC = 80^\circ$
Q5: Cyclic Quadrilateral
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$ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle DBC = 70^\circ$, $\angle BAC = 30^\circ$, find $\angle BCD$. Further, if $AB = BC$, find $\angle ECD$.
$\angle BDC = \angle BAC = 30^\circ$ (Angles in the same segment).
In $\triangle BCD$:
$\angle BCD + \angle DBC + \angle BDC = 180^\circ$.
$\angle BCD + 70^\circ + 30^\circ = 180^\circ$.
$\angle BCD = 180^\circ - 100^\circ = 80^\circ$.
$\angle BCD + \angle DBC + \angle BDC = 180^\circ$.
$\angle BCD + 70^\circ + 30^\circ = 180^\circ$.
$\angle BCD = 180^\circ - 100^\circ = 80^\circ$.
If AB = BC:
In $\triangle ABC$, $\angle BAC = \angle BCA = 30^\circ$ (Angles opposite to equal sides).
$\angle ECD = \angle BCD - \angle BCA$.
$\angle ECD = 80^\circ - 30^\circ = 50^\circ$.
In $\triangle ABC$, $\angle BAC = \angle BCA = 30^\circ$ (Angles opposite to equal sides).
$\angle ECD = \angle BCD - \angle BCA$.
$\angle ECD = 80^\circ - 30^\circ = 50^\circ$.
$\angle BCD = 80^\circ, \angle ECD = 50^\circ$
Q6: Intersecting Chords
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In the figure below, $A, B, C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC = 130^\circ$ and $\angle ECD = 20^\circ$. Find $\angle BAC$.
$\angle BEC + \angle DEC = 180^\circ$ (Linear Pair).
$130^\circ + \angle DEC = 180^\circ \Rightarrow \angle DEC = 50^\circ$.
$130^\circ + \angle DEC = 180^\circ \Rightarrow \angle DEC = 50^\circ$.
In $\triangle DEC$:
$\angle DCE + \angle DEC + \angle CDE = 180^\circ$.
$20^\circ + 50^\circ + \angle CDE = 180^\circ$.
$\angle CDE = 180^\circ - 70^\circ = 110^\circ$.
$\angle DCE + \angle DEC + \angle CDE = 180^\circ$.
$20^\circ + 50^\circ + \angle CDE = 180^\circ$.
$\angle CDE = 180^\circ - 70^\circ = 110^\circ$.
Angles in the same segment are equal.
$\angle BAC = \angle BDC = \angle CDE$.
$\therefore \angle BAC = 110^\circ$.
$\angle BAC = \angle BDC = \angle CDE$.
$\therefore \angle BAC = 110^\circ$.
$\angle BAC = 110^\circ$
Q7: Diagonals as Diameters
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If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Let $ABCD$ be the cyclic quadrilateral. Diagonals $AC$ and $BD$ are diameters.
Since $AC$ is a diameter, angle in a semicircle is $90^\circ$.
$\therefore \angle ABC = 90^\circ$ and $\angle ADC = 90^\circ$.
$\therefore \angle ABC = 90^\circ$ and $\angle ADC = 90^\circ$.
Since $BD$ is a diameter, angle in a semicircle is $90^\circ$.
$\therefore \angle BAD = 90^\circ$ and $\angle BCD = 90^\circ$.
$\therefore \angle BAD = 90^\circ$ and $\angle BCD = 90^\circ$.
Since all interior angles of the quadrilateral are $90^\circ$, $ABCD$ is a rectangle.
Proved.
Q8: Isosceles Trapezium
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If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Given: Trapezium $ABCD$ with $AB \parallel DC$ and $AD = BC$.
Draw $AM \perp DC$ and $BN \perp DC$.
Draw $AM \perp DC$ and $BN \perp DC$.
In $\triangle AMD$ and $\triangle BNC$:
1. $AD = BC$ (Given)
2. $\angle AMD = \angle BNC = 90^\circ$ (Construction)
3. $AM = BN$ (Distance between parallel lines)
$\therefore \triangle AMD \cong \triangle BNC$ (RHS Rule).
$\Rightarrow \angle ADC = \angle BCD$ (CPCT).
1. $AD = BC$ (Given)
2. $\angle AMD = \angle BNC = 90^\circ$ (Construction)
3. $AM = BN$ (Distance between parallel lines)
$\therefore \triangle AMD \cong \triangle BNC$ (RHS Rule).
$\Rightarrow \angle ADC = \angle BCD$ (CPCT).
Since $AB \parallel DC$, $\angle BAD + \angle ADC = 180^\circ$ (Interior angles).
Substitute $\angle ADC$ with $\angle BCD$:
$\angle BAD + \angle BCD = 180^\circ$.
Substitute $\angle ADC$ with $\angle BCD$:
$\angle BAD + \angle BCD = 180^\circ$.
Since the sum of opposite angles is $180^\circ$, quadrilateral $ABCD$ is cyclic.
Proved.
Q9: Intersecting Circles
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Two circles intersect at two points $B$ and $C$. Through $B$, two line segments $ABD$ and $PBQ$ are drawn to intersect the circles at $A, D$ and $P, Q$ respectively. Prove that $\angle ACP = \angle QCD$.
In the first circle (left):
Angles in the same segment are equal.
$\angle ACP = \angle ABP$.
Angles in the same segment are equal.
$\angle ACP = \angle ABP$.
In the second circle (right):
Angles in the same segment are equal.
$\angle QCD = \angle QBD$.
Angles in the same segment are equal.
$\angle QCD = \angle QBD$.
$\angle ABP = \angle QBD$ (Vertically Opposite Angles).
Therefore, $\angle ACP = \angle QCD$.
Proved.
Q10: Circles on Sides
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If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.
Let $\triangle ABC$ be the triangle. Two circles are drawn with $AB$ and $AC$ as diameters.
Let them intersect at $A$ and $D$.
Let them intersect at $A$ and $D$.
Join $AD$.
Since $AB$ is a diameter, $\angle ADB = 90^\circ$ (Angle in a semicircle).
Since $AC$ is a diameter, $\angle ADC = 90^\circ$ (Angle in a semicircle).
Since $AB$ is a diameter, $\angle ADB = 90^\circ$ (Angle in a semicircle).
Since $AC$ is a diameter, $\angle ADC = 90^\circ$ (Angle in a semicircle).
$\angle ADB + \angle ADC = 90^\circ + 90^\circ = 180^\circ$.
Thus, $BDC$ is a straight line.
Thus, $BDC$ is a straight line.
Therefore, $D$ lies on the third side $BC$.
Proved.
Q11: Common Hypotenuse
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$ABC$ and $ADC$ are two right triangles with common hypotenuse $AC$. Prove that $\angle CAD = \angle CBD$.
Since $\triangle ABC$ and $\triangle ADC$ are right-angled at $B$ and $D$ respectively,
$\angle ABC = \angle ADC = 90^\circ$.
$\angle ABC = \angle ADC = 90^\circ$.
$\angle ABC + \angle ADC = 180^\circ$.
Thus, quadrilateral $ABCD$ is cyclic.
Thus, quadrilateral $ABCD$ is cyclic.
Consider the chord $CD$.
Angles in the same segment are equal.
$\angle CAD = \angle CBD$.
Angles in the same segment are equal.
$\angle CAD = \angle CBD$.
Proved.
Q12: Cyclic Parallelogram
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Prove that a cyclic parallelogram is a rectangle.
Let $ABCD$ be a cyclic parallelogram.
Since it is a parallelogram, $\angle A = \angle C$ (Opposite angles are equal).
Since it is a parallelogram, $\angle A = \angle C$ (Opposite angles are equal).
Since it is cyclic, $\angle A + \angle C = 180^\circ$ (Sum of opposite angles).
$\angle A + \angle A = 180^\circ \Rightarrow 2\angle A = 180^\circ \Rightarrow \angle A = 90^\circ$.
A parallelogram with one angle $90^\circ$ is a rectangle.
Proved.