Circles
NCERT Solutions • Class 9 Maths • Chapter 9Exercise 9.3
1. In Fig. 9.23, 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 $\angle ADC$.
Step 1: Calculate the total angle subtended at the centre.
$\angle AOC = \angle AOB + \angle BOC = 60^\circ + 30^\circ = 90^\circ$.
Step 2: Use the theorem that angle at the centre is double the angle at the remaining part of the circle.
$\angle ADC = \frac{1}{2} \angle AOC$
$\angle ADC = \frac{1}{2} \times 90^\circ$
$\angle ADC = 45^\circ$.
$\angle AOC = \angle AOB + \angle BOC = 60^\circ + 30^\circ = 90^\circ$.
Step 2: Use the theorem that angle at the centre is double the angle at the remaining part of the circle.
$\angle ADC = \frac{1}{2} \angle AOC$
$\angle ADC = \frac{1}{2} \times 90^\circ$
$\angle ADC = 45^\circ$.
2. 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.
Step 1: Let the chord be $AB$ and centre $O$.
Since Chord = Radius, $\Delta OAB$ is an equilateral triangle.
Therefore, Angle at centre $\angle AOB = 60^\circ$.
Step 2: Angle on the Major Arc ($x$):
$x = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60^\circ =$ $30^\circ$.
Step 3: Angle on the Minor Arc ($y$):
Points on the major and minor arc form a cyclic quadrilateral.
$x + y = 180^\circ \Rightarrow 30^\circ + y = 180^\circ$
$y = 150^\circ$.
Ans: Major Arc: $30^\circ$, Minor Arc: $150^\circ$.
Since Chord = Radius, $\Delta OAB$ is an equilateral triangle.
Therefore, Angle at centre $\angle AOB = 60^\circ$.
Step 2: Angle on the Major Arc ($x$):
$x = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60^\circ =$ $30^\circ$.
Step 3: Angle on the Minor Arc ($y$):
Points on the major and minor arc form a cyclic quadrilateral.
$x + y = 180^\circ \Rightarrow 30^\circ + y = 180^\circ$
$y = 150^\circ$.
Ans: Major Arc: $30^\circ$, Minor Arc: $150^\circ$.
3. In Fig. 9.24, $\angle PQR = 100^\circ$, where P, Q and R are points on a circle with centre O. Find $\angle OPR$.
Step 1: Find reflex $\angle POR$.
Angle at centre = $2 \times$ Angle at circumference.
Reflex $\angle POR = 2 \times 100^\circ = 200^\circ$.
Step 2: Find internal $\angle POR$.
$\angle POR = 360^\circ – 200^\circ = 160^\circ$.
Step 3: In $\Delta OPR$, $OP = OR$ (Radii).
$\angle OPR = \angle ORP$ (Angles opposite equal sides).
$\angle OPR + \angle ORP + \angle POR = 180^\circ$
$2\angle OPR + 160^\circ = 180^\circ$
$2\angle OPR = 20^\circ \Rightarrow$ $\angle OPR = 10^\circ$.
Angle at centre = $2 \times$ Angle at circumference.
Reflex $\angle POR = 2 \times 100^\circ = 200^\circ$.
Step 2: Find internal $\angle POR$.
$\angle POR = 360^\circ – 200^\circ = 160^\circ$.
Step 3: In $\Delta OPR$, $OP = OR$ (Radii).
$\angle OPR = \angle ORP$ (Angles opposite equal sides).
$\angle OPR + \angle ORP + \angle POR = 180^\circ$
$2\angle OPR + 160^\circ = 180^\circ$
$2\angle OPR = 20^\circ \Rightarrow$ $\angle OPR = 10^\circ$.
4. In Fig. 9.25, $\angle ABC = 69^\circ$, $\angle ACB = 31^\circ$, find $\angle BDC$.
Step 1: In $\Delta ABC$:
$\angle BAC + \angle ABC + \angle ACB = 180^\circ$
$\angle BAC + 69^\circ + 31^\circ = 180^\circ$
$\angle BAC + 100^\circ = 180^\circ \Rightarrow \angle BAC = 80^\circ$.
Step 2: Angles in the same segment are equal.
$\angle BDC = \angle BAC$.
$\angle BDC = 80^\circ$.
$\angle BAC + \angle ABC + \angle ACB = 180^\circ$
$\angle BAC + 69^\circ + 31^\circ = 180^\circ$
$\angle BAC + 100^\circ = 180^\circ \Rightarrow \angle BAC = 80^\circ$.
Step 2: Angles in the same segment are equal.
$\angle BDC = \angle BAC$.
$\angle BDC = 80^\circ$.
5. In Fig. 9.26, 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 $\angle BAC$.
Step 1: Find $\angle CED$.
$\angle BEC + \angle CED = 180^\circ$ (Linear Pair).
$130^\circ + \angle CED = 180^\circ \Rightarrow \angle CED = 50^\circ$.
Step 2: In $\Delta CDE$:
$\angle CDE + \angle CED + \angle ECD = 180^\circ$
$\angle CDE + 50^\circ + 20^\circ = 180^\circ$
$\angle CDE = 110^\circ$.
Step 3: Angles in the same segment are equal.
$\angle BAC = \angle CDE$ (or $\angle BDC$).
$\angle BAC = 110^\circ$.
$\angle BEC + \angle CED = 180^\circ$ (Linear Pair).
$130^\circ + \angle CED = 180^\circ \Rightarrow \angle CED = 50^\circ$.
Step 2: In $\Delta CDE$:
$\angle CDE + \angle CED + \angle ECD = 180^\circ$
$\angle CDE + 50^\circ + 20^\circ = 180^\circ$
$\angle CDE = 110^\circ$.
Step 3: Angles in the same segment are equal.
$\angle BAC = \angle CDE$ (or $\angle BDC$).
$\angle BAC = 110^\circ$.
6. 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 $\angle ECD$.
Step 1: Find $\angle BDC$.
$\angle BDC = \angle BAC = 30^\circ$ (Angles in same segment).
Step 2: Find $\angle BCD$.
In $\Delta BCD$, sum of angles is $180^\circ$.
$\angle DBC + \angle BDC + \angle BCD = 180^\circ$
$70^\circ + 30^\circ + \angle BCD = 180^\circ$
$\angle BCD = 80^\circ$.
$\angle BDC = \angle BAC = 30^\circ$ (Angles in same segment).
Step 2: Find $\angle BCD$.
In $\Delta BCD$, sum of angles is $180^\circ$.
$\angle DBC + \angle BDC + \angle BCD = 180^\circ$
$70^\circ + 30^\circ + \angle BCD = 180^\circ$
$\angle BCD = 80^\circ$.
Step 3: If $AB = BC$.
In $\Delta ABC$, angles opposite equal sides are equal $\Rightarrow \angle BCA = \angle BAC = 30^\circ$.
$\angle ECD = \angle BCD – \angle BCA$
$\angle ECD = 80^\circ – 30^\circ$
$\angle ECD = 50^\circ$.
In $\Delta ABC$, angles opposite equal sides are equal $\Rightarrow \angle BCA = \angle BAC = 30^\circ$.
$\angle ECD = \angle BCD – \angle BCA$
$\angle ECD = 80^\circ – 30^\circ$
$\angle ECD = 50^\circ$.
7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Proof:
Let ABCD be the cyclic quadrilateral and AC, BD be diagonals which are also diameters.
Since AC is a diameter, angle in a semi-circle is $90^\circ$.
$\therefore \angle ABC = 90^\circ$ and $\angle ADC = 90^\circ$.
Similarly, BD is a diameter $\Rightarrow \angle DAB = 90^\circ$ and $\angle BCD = 90^\circ$.
Since all interior angles are $90^\circ$, ABCD is a rectangle.
Let ABCD be the cyclic quadrilateral and AC, BD be diagonals which are also diameters.
Since AC is a diameter, angle in a semi-circle is $90^\circ$.
$\therefore \angle ABC = 90^\circ$ and $\angle ADC = 90^\circ$.
Similarly, BD is a diameter $\Rightarrow \angle DAB = 90^\circ$ and $\angle BCD = 90^\circ$.
Since all interior angles are $90^\circ$, ABCD is a rectangle.
8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Given: Trapezium ABCD with $AB \parallel DC$ and $AD = BC$.
Construction: Draw $AM \perp DC$ and $BN \perp DC$.
Proof: In $\Delta AMD$ and $\Delta BNC$:
1. $AD = BC$ (Given)
2. $AM = BN$ (Distance between parallel lines)
3. $\angle AMD = \angle BNC = 90^\circ$
$\therefore \Delta AMD \cong \Delta BNC$ (RHS) $\Rightarrow \angle D = \angle C$.
Since $AB \parallel DC$, interior angles $\angle A + \angle D = 180^\circ$.
Substitute $\angle D$ with $\angle C$: $\angle A + \angle C = 180^\circ$.
Since sum of opposite angles is $180^\circ$, ABCD is cyclic.
Construction: Draw $AM \perp DC$ and $BN \perp DC$.
Proof: In $\Delta AMD$ and $\Delta BNC$:
1. $AD = BC$ (Given)
2. $AM = BN$ (Distance between parallel lines)
3. $\angle AMD = \angle BNC = 90^\circ$
$\therefore \Delta AMD \cong \Delta BNC$ (RHS) $\Rightarrow \angle D = \angle C$.
Since $AB \parallel DC$, interior angles $\angle A + \angle D = 180^\circ$.
Substitute $\angle D$ with $\angle C$: $\angle A + \angle C = 180^\circ$.
Since sum of opposite angles is $180^\circ$, ABCD is cyclic.
9. 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 $\angle ACP = \angle QCD$.
Proof:
1. For the first circle, angles in the same segment are equal:
$\angle ACP = \angle ABP$.
2. For the second circle, angles in the same segment are equal:
$\angle QCD = \angle QBD$.
3. $\angle ABP = \angle QBD$ (Vertically Opposite Angles).
From 1, 2 and 3:
$\angle ACP = \angle QCD$.
1. For the first circle, angles in the same segment are equal:
$\angle ACP = \angle ABP$.
2. For the second circle, angles in the same segment are equal:
$\angle QCD = \angle QBD$.
3. $\angle ABP = \angle QBD$ (Vertically Opposite Angles).
From 1, 2 and 3:
$\angle ACP = \angle QCD$.
10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Proof: Let $\Delta ABC$ have sides AB and AC as diameters. Let circles intersect at A and D.
Join AD.
Since AB is a diameter, $\angle ADB = 90^\circ$ (Angle in semi-circle).
Since AC is a diameter, $\angle ADC = 90^\circ$ (Angle in semi-circle).
Adding both: $\angle ADB + \angle ADC = 90^\circ + 90^\circ = 180^\circ$.
Since $\angle BDC = 180^\circ$, BDC is a straight line.
Therefore, D lies on the third side BC.
Join AD.
Since AB is a diameter, $\angle ADB = 90^\circ$ (Angle in semi-circle).
Since AC is a diameter, $\angle ADC = 90^\circ$ (Angle in semi-circle).
Adding both: $\angle ADB + \angle ADC = 90^\circ + 90^\circ = 180^\circ$.
Since $\angle BDC = 180^\circ$, BDC is a straight line.
Therefore, D lies on the third side BC.
11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle CAD = \angle CBD$.
Proof:
Since $\angle B = 90^\circ$ and $\angle D = 90^\circ$, $\angle B + \angle D = 180^\circ$.
Thus, ABCD is a cyclic quadrilateral (Sum of opposite angles is 180).
Consider the chord CD. Angles in the same segment are equal.
$\therefore$ $\angle CAD = \angle CBD$.
Since $\angle B = 90^\circ$ and $\angle D = 90^\circ$, $\angle B + \angle D = 180^\circ$.
Thus, ABCD is a cyclic quadrilateral (Sum of opposite angles is 180).
Consider the chord CD. Angles in the same segment are equal.
$\therefore$ $\angle CAD = \angle CBD$.
12. Prove that a cyclic parallelogram is a rectangle.
Proof: Let ABCD be a cyclic parallelogram.
1. Since it is a parallelogram, opposite angles are equal: $\angle A = \angle C$.
2. Since it is cyclic, opposite angles sum to $180^\circ$: $\angle A + \angle C = 180^\circ$.
From 1 and 2:
$2\angle A = 180^\circ \Rightarrow \angle A = 90^\circ$.
A parallelogram with one angle $90^\circ$ is a rectangle.
1. Since it is a parallelogram, opposite angles are equal: $\angle A = \angle C$.
2. Since it is cyclic, opposite angles sum to $180^\circ$: $\angle A + \angle C = 180^\circ$.
From 1 and 2:
$2\angle A = 180^\circ \Rightarrow \angle A = 90^\circ$.
A parallelogram with one angle $90^\circ$ is a rectangle.