# Exercise E10.5 Circles NCERT Solutions Class 9

## Chapter 10 Ex.10.5 Question 1

In the given figure, \(A,\, B\) and \(C\) are three points on a circle with center \(O\) such that \(\begin{align}\angle {BOC}=30^{\circ} \text { and } \angle {AOB}=60^{\circ} \end {align}\)If \(D\) is a point on the circle other than the arc \(ABC,\) find \(\begin{align} \angle {ADC} \end{align}\)

**Solution**

**Video Solution**

**What is known?**

Two angles subtended by arcs at the centre.

**What is unknown?**

Value of \( \angle {ADC.}\)

**Reasoning:**

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.

**Steps:**

\[\begin{align} \angle {AOC} &=\angle {AOB}+\angle {BOC} \\ &=90^\circ \end{align}\]

By Theorem \(10.8\) ,

\[\begin{align} \angle {AOC} &=2 \angle {ADC} \\ \angle ADC &= \frac{1}{2}\angle AOC\\\angle ADC &= \frac{1}{2} \times 90 = \,45^\circ \\\therefore \angle ADC &= \,45^\circ \end{align}\]

## Chapter 10 Ex.10.5 Question 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.

**Solution**

**Video Solution**

**What is known?**

Chord’s length is equal to the radius.

**What is unknown?**

Angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

**Reasoning:**

- 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.
- A quadrilateral \(ABCD\) is called cyclic if all the four vertices of it lie on a circle.
- The sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin{align}180^{\circ}.\end {align}\)

**Steps:**

Draw a circle with any radius and centre \(O\). Let \(AO\) and \(BO\) be the **\(2\)** radii of the circle and let \(AB\) be the chord equal to the length of radius. Join them to form a triangle.

\(OA = OB = AB\)

Hence \(\begin{align} \Delta {ABO} \end {align}\) becomes an equilateral triangle.

Draw **\(2\)** points \(C\) and \(D\) on the circle such that they lie on major arc and minor arc respectively.

Since \(\begin{align} \Delta {ABO} \end {align}\) is an equilateral triangle, we get \(\begin{align} \angle {ABO}=60^{\circ} \end {align}\)

For the arc \(AB,\) \(\begin {align} \angle {AOB}=2 \angle {ACB} \end {align}\) as we know that 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.

\[\begin{align}\angle ACB &= \frac{1}{2}\angle AOB\\&= \frac{1}{2} \times 60 \\&= 30^\circ \end{align}\]

As you can notice the points \(A, B, C\) and \(D\) lie on the circle. Hence \(A B C D\) is a cyclic quadrilateral.

We know that, the sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin{align}180^\circ.\end {align}\)

Therefore,

\[\begin{align}\angle ACB + \angle ADB &= 180^\circ \\30 + \angle ADB &= 180\\\angle ADB &= 150^\circ \end{align}\]

So when the chord of a circle is equal to the radius of the circle, the angle subtended by the chord at a point on the minor arc is \(\begin{align}150^\circ.\end {align}\) and also at a point on the major arc is \(\begin{align}30^\circ.\end {align}\)

## Chapter 10 Ex.10.5 Question 3

In the given figure, \(\begin {align} \angle {PQR}=100^{\circ} \end {align}\) where \(P, Q\) and \(R\) are points on a circle with center \(O\) . Find \(\begin {align}\angle {OPR} \end {align}\).

**Solution**

**Video Solution**

**What is known?**

**\(3\)** points on the circle and

\(\begin {align}\angle {PQR}=100^{\circ} \end {align}\)

**What is unknown?**

Value of \(\angle {PQR} \)

**Reasoning:**

- 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.
- A quadrilateral \(\begin {align} {ABCD} \end {align}\) is called cyclic if all the four vertices of it lie on a circle.
- The sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin {align} 180^{\circ}. \end {align}\)

**Steps:**

Mark any point on the major arc side (opposite side to point \(\begin {align} {Q}\end {align}\)) as \(\begin {align} {S.}\end {align}\)

Since all points \(\begin {align} {P, Q, R, S}\end {align}\) lie on the circle, \(\begin {align} {PQRS}\end {align}\) becomes a cyclic quadrilateral.

We know that, the sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin {align} 180^\circ.\end {align}\)

Therefore,

\[\begin{align} \angle {PQR}+\angle {PSR} &=180^{\circ} \\ 100^{\circ}+\angle {PSR} &=180^{\circ} \\ \angle {PSR} &=180^{\circ}-100^{\circ} \\ &=80^{\circ} \end{align}\]

We know that, 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,

\[\begin{aligned} \angle {POR} &=2 \angle {PSR} \\ &=2 \times 80^{\circ} \\ &=160^{\circ} \end{aligned}\]

Consider the \(\begin{aligned} \Delta {OPR.}\end{aligned}\) It is an isosceles triangle as

\[\begin{align} OP &= OR \\ &= \text{Radius of the circle.}\end{align}\]

\[\begin {align}∴ \angle {OPR}=\angle {ORP} \end {align}\]

Sum of all angles in a triangle is \(\begin {align}180^{\circ}. \end {align}\)

Therefore,

\[\begin{align}\angle {OPR}+ \!\angle {POR}+ \! \angle {ORP}&=180^{\circ} \\ \angle {OPR}+160^{\circ}+\angle {OPR}&=180^{\circ} \end{align}\]

\[\begin{align} 2 \angle {OPR}&= 180^{\circ} -160^{\circ} \\ \angle {OPR}&=10^{0}\end{align}\]

## Chapter 10 Ex.10.5 Question 4

In the given figure, \(\begin {align} \angle {ABC}=69^{\circ} \end {align}\) and \(\begin {align} \angle {ACB}=31^{\circ} \end {align}\)find \(\begin {align} \angle {BDC} . \end {align}\)

**Solution**

**Video Solution**

**What is known?**

Two angles in a triangle.

**What is unknown?**

Value of \(\begin {align} \angle {BDC}\end {align}\)

**Reasoning:**

- Sum of angles in a triangle is \(\begin {align} 180^{\circ}.\end {align}\)
- Angles in the same segment are equal.

**Steps:**

Consider the \(\begin {align} \Delta {ABC,}\end {align}\) the sum of all angles will be \(\begin {align} 180^\circ.\end {align}\)

\[\begin{align} \angle {ABC}+ \!\!\angle {BAC}+ \!\angle {ACB} &=180^{\circ} \\ 69^{\circ}+\angle {BAC}+31^{\circ} &=180^{\circ} \end{align}\]

\[\begin{align} \angle {BAC} &=180^{\circ}-\left(69^{\circ}+31^{\circ}\right) \\ &=180^{\circ}-100^{\circ} \\ &=80^{\circ} \end{align}\]

We know that, angles in the same segment of a circle are equal.

\[\begin {align} ∴ \angle {BDC}=\angle {BAC}=80^{\circ} \end {align}\]

## Chapter 10 Ex.10.5 Question 5

In the given figure, \(A, B, C\) and \(D\) are four points on a circle. \(AC\) and \(BD\) intersect at a point \(\begin {align}{E} \end {align}\) such that \(\begin {align} \angle {BEC}=130^{\circ} \end {align}\) and \(\begin {align} \angle {ECD}=20^{\circ}. \end {align}\) Find \(\begin {align} \angle {BAC}\end {align}\).

**Solution**

**Video Solution**

**What is known?**

Two angles \(\begin {align} \angle {BEC}=130^{\circ} \end {align}\)

and \(\begin {align} \angle {ECD}=20^{\circ} \end {align}\)

**What is unknown?**

Value of \( \angle {BAC} \)

**Reasoning:**

- Sum of angles in a triangle is \(180^{\circ}.\)
- Angles in the same segment are equal.

**Steps:**

Consider the straight line \(\begin {align} {BD} \end {align}\). As the line \(\begin {align} {AC} \end {align}\) intersects with the line \(\begin {align} {BD,} \end {align}\) then the sum of two adjacent angles so formed is \(180^{\circ}.\)

Therefore,

\[\begin{align} \angle {BEC}+\angle {DEC} &=180^{\circ} \\ 130^{\circ}+\angle {DEC} &=180^{\circ} \\ \angle {DEC} &=180^{\circ}-130^{\circ} \\ &=50^{\circ} \end{align}\]

Consider the \(\begin{align} \Delta {DEC,} \end{align}\) the sum of all angles will be \(180^{\circ}.\)

\[\begin{align} \angle {DEC}+\angle {EDC}+\angle {ECD} &=180^{\circ} \\ 50^{\circ}+\angle {EDC}+20^{\circ} &=180^{\circ} \end{align}\]

\[\begin{align} \angle {EDC} &=180^{\circ}-70^{\circ} \\ &=110^{\circ} \\ ∴ \angle {BDC}=\angle {EDC} &=110^{\circ} \end{align}\]

We know that, angles in the same segment of a circle are equal.

\[\begin {align} ∴ \angle {BAC}=\angle {BDC}=110^{\circ}\end {align}\]

## Chapter 10 Ex.10.5 Question 6

\(\begin {align} {ABCD}\end {align}\) is a cyclic quadrilateral whose diagonals intersect at a point \(\begin {align} {E. }\end {align}\)If \(\begin {align} \angle {DBC}=70^{\circ}, \quad \angle {BAC}=30^{\circ} \end {align}\) find \(\begin {align} \angle {BCD}. \end {align}\) Further if \(\begin {align} {AB=BC,} \end {align}\) find \(\begin {align} \angle {ECD.} \end {align}\)

**Solution**

**Video Solution**

**What is known?**

\(\begin {align} {ABCD}\end {align}\) is cyclic quadrilateral. \(\begin {align} {DBC}=70^\circ \end {align}\) and \(\begin {align} {BAC}=30^\circ. \end {align}\) \(\begin {align}{AB = BC.}\end {align}\)

**What is unknown?**

Value of \(\begin {align} \angle {BCD} \text { and } \angle {ECD} \end {align}\)

**Reasoning:**

- A quadrilateral \(\begin {align} {ABCD}\end {align}\) is called cyclic if all the four vertices of it lie on a circle.
- The sum of either pair of opposite angles of a cyclic quadrilateral is \(180^\circ.\)
- Sum of angles in a triangle is \(180^\circ.\)
- Angles in the same segment are equal.

**Steps:**

Based on the data given, draw the figure.

In the triangles \({ABD} \) and \({BCD,} \) \(\begin {align} \angle {CAD}=\angle {CBD}=70^{\circ}. \end {align}\) (Angles in the same segment)

So \(\begin {align} \angle {BAD}=30^{\circ}+70^{\circ}=100^{\circ} \end {align}\)

Since \(\begin {align} {ABCD }\end {align}\)is a cyclic quadrilateral, the sum of either pair of opposite angles of cyclic quadrilateral is \(180^{\circ}. \)

\[\begin{align} \angle {BAD}+ \angle {BCD}&=180^{\circ} \\ \angle {BCD} &=180^{\circ}-100^{\circ} \\&=80^{\circ} \end{align}\]

Also given \(\begin{align} {AB = BC.}\end{align}\)

So, \(\begin {align} \angle {BCA}=\angle {BAC}=30^{\circ} \end {align}\) (Base angles of isosceles triangle are equal)

\[\begin{align} ∴ \quad \angle {ECD} &=\angle {BCD}-\angle {BCA} \\ &=80^{\circ}-30^{\circ} \\ &=50^{\circ} \end{align}\]

## Chapter 10 Ex.10.5 Question 7

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

**Solution**

**Video Solution**

**What is known?**

Diagonals of a cyclic quadrilateral are diameters of a circle passing through the vertices.

**What is unknown?**

To prove lines joining the vertices is a rectangle.

**Reasoning:**

- The sum of either pair of opposite angles of a cyclic quadrilateral is \(180^{\circ}.\)
- Diameter is a chord.

**Steps:**

Let \({DB}\) be the diameter of the circle which is also a chord.

Then \(\begin {align} \angle {BOD}=180^{\circ} \end {align}\)

We know that, 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.

\[\begin{align} \angle {BAD} &=\frac{1}{2} \times \angle {BOD} \\ &=90^{\circ} \end{align}\]

Similarly, \(\begin {align} \angle {BCD}=90^{\circ} \end {align}\)

Now considering \(\text{AC}\) as the diameter of the circle, we get \(\begin {align} \angle {AOC}=180^{\circ} \end {align}\)

We know that, 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.

\[\begin{align} \angle {ABC} &=\frac{1}{2} \times \angle {AOC} \\ &=90^{\circ} \end{align}\]

Similarly, \(\begin {align} \angle {ADC}=90^{\circ} \end {align}\)

As you can see, all the angles at the corners are \(90^\circ \)we can say that the shape joining the vertices is a rectangle.

This problem can also be solved by using the property of cyclic quadrilaterals.

## Chapter 10 Ex.10.5 Question 8

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

**Solution**

**Video Solution**

**What is known?**

Non-parallel sides of trapezium are equal.

**What is unknown?**

To prove that trapezium is cyclic.

**Reasoning:**

If the sum of a pair of opposite angles of a quadrilateral is \(180^{\circ},\) the quadrilateral iscyclic. Using Right angled-Hypotenuse-Side (**RHS**) criteria and Corresponding parts of congruent triangles (**CPCT**) we prove the statement.

**Steps:**

Draw a trapezium \( {ABCD} \)

with \( {AB} \| {CD} \)

\(AD\) and \(BC\) are the non-parallel sides which are equal. \({AD = BC.}\)

Draw \(\begin {align} {AM} \perp {CD} \end {align}\)

and \(\begin {align} {BN} \perp {CD} \end {align}\)

Consider \(\Delta {AMD}\) and \(\Delta \text{BNC.}\)

\(\begin{align} {AD} &={BC} \qquad \; (\text { Given }) \\\angle{AMD}&=\angle {BNC} \quad \left(90^{\circ}\right) \end{align}\)

\(AM =BN \) (Perpendicular distance between two parallel lines is same)

By **RHS** congruence,

\[\begin {align}\Delta {AMD} \cong \Delta {BNC} \end {align}\]

Using CPCT ,

\[\begin {align} \angle {ADC}=\angle {BCD} \ldots . .(1) \end {align}\]

\(\angle {BAD} \) and \(\angle {ADC} \) are on the same side of transversal \(\angle \text{AD}. \)

\[\begin{align}\angle {BAD}+\angle {ADC}&=180^{\circ} \\ \angle {BAD}+\angle {BCD}&=180^{\circ} \text{[Using}(1)]\end{align}\]

This equation proves that the opposite angles are supplementary.

Hence, \({ABCD} \) is a cyclic quadrilateral.

## Chapter 10 Ex.10.5 Question 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 \(A, D, P\) and \(Q\) respectively. Prove that \(\begin {align} \angle {ACP}=\angle {QCD.} \end {align}\)

** **

** **

**Solution**

**Video Solution**

**What is given ?**

Two circles intersect at two points

**What is unknown?**

Proof of \( \angle {ACP}=\angle {QCD} \)

**Reasoning:**

\(\angle {ACP} \) and \(\angle {ABP} \) lie on the same segment.

Similarly, \( \angle {DCQ} \) and \( \angle {DBQ} \) lie on the same segment.

Angles in the same segment of a circle are equal.

**Steps:**

We know that, angles in the same segment of a circle are equal.

So we get \( \angle {ACP}=\angle {ABP} \) and \( \angle {QCD}=\angle {QBD} \)

Also, \(\begin {align} \angle {QBD}=\angle {ABP} \end {align}\)(Vertically opposite angles)

Therefore \( \angle {ACP}=\angle {QCD} \)

## Chapter 10 Ex.10.5 Question 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.

**Solution**

**Video Solution**

**What is given ?**

Two circles are drawn taking two sides of a triangle as diameters.

**What is unknown?**

To prove that point of intersection of the **\(2\)** circles lie on the third side.

**Reasoning:**

Angle in a semicircle is a right angle. By using this fact we can show that \(BDC\) is a line which will lead to the proof that point of intersection lie on the third side.

**Steps:**

Since angle in a semicircle is a right angle, we get:

\( \begin{align} \angle {ADB}&=90^{\circ} \text { and } \angle {ADC}=90^{\circ} \end{align}\)

\(\begin{align} \angle {ADB}+\angle {ADC}&=90^{\circ}+90^{\circ} \\ \Rightarrow \;\; \angle {ADB}+\angle {ADC}&=180^{\circ} \end{align}\)

\(\Rightarrow \quad{BDC} \text { is a straight line. }\)

∴ \({D}\) lies on \({BC}\)

Hence, point of intersection of circles lie on the third side \(BC.\)

## Chapter 10 Ex.10.5 Question 11

\({ABC}\) and \({ADC}\) are two right triangles with common hypotenuse \({AC.}\) Prove that \(\angle{CAD}= \angle {CBD}.\)

**Solution**

**Video Solution**

**What is given ?**

\({ABC}\) and \({ADC}\) are two right triangles with common hypotenuse \({AC.}\)

**What is unknown?**

Proof of \(\angle{CAD}= \angle {CBD}\)

**Reasoning:**

Sum of all angles in a triangle is \(180^\circ\)

If the sum of pair of opposite angles in a quadrilateral is \(180\) then it is cyclic quarilateral

Angles in the same segment of a circle are equal.

**Steps:**

Consider \(\Delta {ABC,}\)

\(\begin{align}&\angle {ABC}+ \angle {BCA}+\angle {CAB} =180^{\circ} \\&\text { (Angle sum property of a triangle) }\end{align}\)

\(\begin{align} 90^{\circ}+\angle {BCA}+\angle {CAB}&=180^{\circ} \\ \angle {BCA}+\angle {CAB}&=90^{\circ} \ldots(1)\end{align}\)

Consider \(\Delta {ADC,}\)

\(\begin{align}&\angle {CDA}+ \angle {ACD}+\angle {DAC} =180^{\circ} \\&\text { (Angle sum property of a triangle) }\end{align}\)

\(\begin {align} 90^{\circ}+\angle {ACD}+\angle {DAC}&=180^{\circ}\\\angle {ACD}+\angle {DAC}&=90^{\circ} \ldots(2) \end {align}\)

Adding Equations (**\(1\)**) and (**\(2\)**), we obtain

\( \begin {align} \begin{Bmatrix} \angle {BCA}+\angle {CAB}\\+\angle {ACD}+\angle {DAC} \end{Bmatrix}&=180^{\circ} \\ \begin{Bmatrix} (\angle {BCA}+\angle {ACD})+ \\ (\angle {CAB}+\angle {DAC}) \end{Bmatrix} &=180^{\circ} \end {align}\)

\( \begin {align} \angle {BCD}+\angle {DAB}&=180^{\circ} \ldots(3) \end {align}\)

However, it is given that

\(\begin {align}\angle {B}+\angle {D} &=90^{\circ}+90^{\circ} \\ &=180^{\circ} \ldots(4) \end {align}\)

From Equations (**\(3\)**) and (**\(4\)**), it can be observed that the sum of the measures of opposite angles of quadrilateral \(\text{ABCD}\) is \(180^{\circ}.\) Therefore, it is a cyclic quadrilateral.

Consider chord \({CD.}\)

\(\begin {align}\angle {CAD} = \angle {CBD} \end {align}\) (Angles in the same segment)

## Chapter 10 Ex.10.5 Question 12

Prove that cyclic parallelogram is a rectangle.

**Solution**

**Video Solution**

**What is given ?**

Cyclic quadrilateral is a parallelogram.

**What is unknown?**

Prove that cyclic parallelogram is a rectangle.

**Reasoning:**

The sum of either pair of opposite angles of a cyclic quadrilateral is \(180^\circ.\) By using this fact we can show each angle of cyclic parallelogram as \(90^\circ\) which will prove the statement it is a rectangle.

**Steps:**

Let \({ABCD}\) be the cyclic parallelogram.

We know that opposite angles of a parallelogram are equal.

\[\begin {align}\angle {A}=\angle {C} \text { and } \angle {B}=\angle {D} \ldots .(1) \end {align}\]

We know that the sum of either pair of opposite angles of a cyclic quadrilateral is \(180^{\circ}.\)

\[\begin {align} \angle {A}+\angle {C}=180^{\circ} \ldots .(2) \end {align}\]

Substituting (**\(1\)**) in (**\(2\)**),

\[\begin{align}\angle {A}+\angle {C}&=180^{\circ} \\ \angle {A}+\angle {A}&=180^{\circ} \\ 2 \angle {A}&=180^{\circ} \\ \angle {A}&=90^{\circ}\end{align}\]

We know that if one of the interior angles of a parallelogram is \(90^{\circ},\) all the other angles will also be equal to \(90^{\circ}.\)

Since all the angles in the parallelogram is \(90^{\circ},\) we can say that parallelogram \({ABCD}\) is a rectangle.