Class 9 - Maths - Circles

**Exercise 10.1**

**Question 1:**

Fill in the blanks:

(i) The centre of a circle lies in_______ of the circle. (exterior/ interior)

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in______ of the circle. (exterior/ interior)

(iii) The longest chord of a circle is a_______ of the circle.

(iv) An arc is a_____ when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and_______ of the circle.

(vi) A circle divides the plane, on which it lies, in______ parts.

Answer:

(i) The centre of a circle lies in ** interior** of the circle.

(ii) A point, whose distance from the centre of a circle is greater than its radius lies in __exterior__

of the circle.

(iii) The longest chord of a circle is a ** diameter** of the circle.

(iv) An arc is a ** semi-circle** when its ends are the ends of a diameter.

(v) Segment of a circle is the region between an arc and ** chord** of the circle.

(vi) A circle divides the plane, on which it lies, in ** three** parts.

**Question 2:**

Write True or False: Give reasons for your answers.

(i) Line segment joining the centre to any point on the circle is a radius of the circle.

(ii) A circle has only finite number of equal chords.

(iii) If a circle is divided into three equal arcs, each is a major arc.

(iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

(v) Sector is the region between the chord and its corresponding arc.

(vi) A circle is a plane figure.

Answer:

(i) True.

All the points on the circle are at equal distance from the center of the circle and this equal

distance is called as radius of the circle.

(ii) False.

There are infinite points on a circle. Therefore, we can draw infinite number of chords of given

length. Hence a circle has infinite number of equal chords.

(iii) False.

Let there are three arcs of same length as AB, BC and CA. It can be observed that the minor arc

is BDC and CAB is a major arc. Therefore, AB, BC and CA are minor arcs of the circle.

(iv) True.

Let AB be a chord which is twice as long as its radius. It can be observed that in this situation,

our chord will be passing through the center of the circle. Therefore, it will be the diameter of

the circle.

(v) False.

Sector is the region between an arc and two radii joining the center to the end points

of the arc. For example, in the given figure, OAB is the sector of the circle.

(vi) True.

A circle is a two-dimensional figure and it can also be referred to as a plane figure.

**Exercise 10.2**

**Question 1:**

Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer:

Given : Two congruent circles with centres

O and O′. AB and CD are equal chords

of the circles with centres O and O′ respectively.

To Prove: ∠AOB = ∠COD

Proof: In triangles AOB and COD,

AB = CD [Given]

AO = CO’ [Radii of congruent circle]

BO = DO’ [Radii of congruent circle]

∆AOB ≅ ∆CO′D [by SSS axiom]

Hence, ∠AOB ≅ ∠CO′D [by CPCT]

**Question 2:**

Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer:

Given : Two congruent circles with

centres O and O′. AB and CD are

chords of circles with centre O

and O′ respectively such that ∠AOB = ∠CO′D

To Prove: AB = CD

Proof: In triangles AOB and CO′D,

AO = CO’ [Radii of congruent circle]

BO = DO’ [Radii of congruent circle]

∠AOB = ∠CO′D [Given]

∆AOB ≅ ∆CO′D [by SAS axiom]

Hence, AB = CD [by CPCT]

**Exercise 10.3**

**Question 1:**

Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Answer:

No common point One common point Two common points

Hence, the maximum number of common point is two.

**Question 2:**

Suppose you are given a circle. Give a construction to find its centre.

Answer:

Steps of Construction:

- Take arc PQ of the given circle.
- Take a point R on the arc PQ and draw chords PR and RQ.
- Draw perpendicular bisectors of PR and RQ. These

perpendicular bisectors intersect at point O.

Hence, point O is the centre of the given circle.

**Question 3:**

If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Answer:

Given: AB is the common chord of two intersecting circles (O, r) and (O′, r′).

To Prove: Centres of both circles lie on the perpendicular bisector of chord AB, i.e., AB is

bisected at right angle by OO′.

Construction: Join AO, BO, AO′ and BO′.

Proof: In ∆AOO′ and ∆BOO′

AO = OB [Radii of the circle (O, r]

AO′ = BO′ [Radii of the circle (O′, r′)]

OO′ = OO′ [Common]

So, ∆AOO′ ≅ ∆BOO′ [by SSS congruency]

Hence, ∠AOO′ = ∠BOO′ [by CPCT]

Now in ∆AOC and ∆BOC,

∠AOC = ∠BOC [∠AOO′ = ∠BOO′]

AO = BO [Radii of the circle (O, r)]

OC = OC [Common]

So, ∆AOC ≅ ∆BOC [by SAS congruency]

Hence, AC = BC and ∠ACO = ∠BCO .............(i) [by CPCT]

=> ∠ACO + ∠BCO = 180^{0} ……….(ii) [Linear pair]

=> ∠ACO = ∠BCO = 90^{0} [From equation 1 and 2]

Hence, OO’ lies on the perpendicular bisector of AB.

**Exercise 10.4**

**Question 1:**

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.

Answer:

In ∆AOO′,

AO^{2} = 5^{2} = 25

AO′^{2} = 3^{2} = 9

OO′^{2} = 4^{2} = 16

AO′^{2} + OO′^{2} = 9 + 16 = 25 = AO^{2}

=> ∠AO′O = 90^{0} [By converse of Pythagoras theorem]

Similarly, ∠BO′O = 90^{0}.

=> ∠AO′B = 90^{0} + 90^{0} = 180^{0}

=> AO′B is a straight line whose mid-point is O.

=> AB = 3 + 3 = 6 cm.

**Question 2:**

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Answer:

Given: AB and CD are two equal chords of a circle which meet at E.

To prove: AE = CE and BE = DE

Construction: Draw OM ⊥ AB and ON ⊥ CD and join OE.

Proof: In ∆OME and ∆ONE,

OM = ON [Equal chords are equidistant]

OE = OE [Common]

∠OME = ∠ONE [Each equal to 90°]

So, ∆OME ≅ ∆ONE [by RHS axiom]

=> EM = EN ........(i) [by CPCT]

Now AB = CD [Given]

=> AB/2 = CD/2

=> AM = CN ………(ii) [Perpendicular from centre bisects the chord]

Adding equation (i) and (ii), we get

EM + AM = EN + CN

=> AE = CE …………….(iii)

Now, AB = CD ……….(iv)

=> AB – AE = CD – AE [From equation (iii)]

=> BE = CD – CE

**Question 3:**

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.

Answer:

Given: AB and CD are two equal chords of a circle which meet at E within the circle and a line

PQ joining the point of intersection to the centre.

To Prove: ∠AEQ = ∠DEQ

Construction: Draw OL ⊥ AB and OM ⊥ CD.

Proof: In ∆OLE and ∆OME, we have

OL = OM [Equal chords are equidistant]

OE = OE [Common]

∠OLE = ∠OME [Since each = 90^{0}]

So, ∆OLE ≅ ∆OME [by RHS congruence]

=> ∠LEO = ∠MEO [CPCT]

**Question 4:**

If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Answer:

Given: A line AD intersects two concentric circles at A, B, C and D, where O is the centre of these circles.

To prove: AB = CD

Construction: Draw OM ⊥ AD.

Proof: AD is the chord of larger circle.

So, AM = DM ………(i) [OM bisects the chord]

BC is the chord of smaller circle

So, BM = CM ……..(ii) [OM bisects the chord]

Subtracting equation (ii) from equation (i), we get

AM – BM = DM – CM

=> AB = CD

**Question 5:**

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park.

Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma.

If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?

Answer:

Let Reshma, Salma and Mandip be represented by R, S and M respectively.

Now, Draw OL ⊥ RS.

OL^{2} = OR^{2} – RL^{2}

=> OL^{2} = 5^{2} – 3^{2} [RL = 3 m, because OL ⊥ RS]

=> OL^{2} = 25 – 9

=> OL^{2} = 16

=> OL = √16

=> OL = 4

Now, area of triangle ORS = (1/2) * KR * OR

= (1/2) * KR * 5

Also, area of ∆ORS = (1/2) * RS * OL

= (1/2) * 6 * 4

= 12 m^{2}

Now, (1/2) * KR * 5 = 12

=> KR = (12 * 2)/5

=> KR = 24/5

=> KR = 4.8 m

Now, RM = 2KR

=> RM = 2 * 4.8

=> RM = 9.6 m

Hence, distance between Reshma and Mandip is 9.6 m.

**Question 6:**

A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each

having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer:

Let Ankur, Syed and David be represented by A, S and D respectively.

Let PD = SP = SQ = QA = AR = RD = x

In ∆OPD,

OP^{2} = 400 – x^{2}

=> OP = √(400 – x^{2})

=> AP = 2√(400 – x^{2}) + √(400 – x^{2}) [Since centroid divides the median in the ratio 2 : 1]

=> AP = 3√(400 – x^{2})

Now, in ∆APD,

PD^{2} = AD^{2} – DP^{2}

=> x^{2} = (2x)^{2} – {3√(400 – x^{2})}^{2}

=> x^{2} = 4x^{2} – 9(400 – x^{2})

=> x^{2} = 4x^{2} – 3600 + 9x^{2}

=> x^{2} = 13x^{2} – 3600

=> 13x^{2} – x^{2} = 3600

=> 12x^{2} = 3600

=> x^{2} = 3600/12

=> x^{2} = 300

=> x = √300

=> x = 10√2

Now, SD = 2x = 2 * 10√3 = 20√3

Hence, ASD is an equilateral triangle.

So, SD = AS = AD = 20√3

Hence, the length of the string of each phone is 20√3 m.

**Exercise 10.5**

**Question 1:**

In Fig. 10.36, A,B and C are three points on a circle with centre O such that ∠ BOC = 30^{0} and ∠ AOB = 60°.

If D is a point on the circle other than the arc ABC, find ∠ADC.

Answer:

We have, ∠BOC = 30^{0} and ∠AOB = 60^{0}

Now, ∠AOC = ∠AOB + ∠BOC

= 60^{0} + 30^{0}

= 90^{0}

We know that angle subtended by an arc at the centre of a circle is double the angle

subtended by the same arc on the remaining part of the circle.

So, 2∠ADC = ∠AOC

=> ∠ADC = ∠AOC/2

= 90^{0}/2

=> ∠ADC = 45^{0}

**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.

Answer:

We have, OA = OB = AB

Therefore, ∆OAB is a equilateral triangle.

=> ∠AOB = 60^{0}

We know that angle subtended by an arc at the centre of a circle is double the angle

subtended by the same arc on the remaining part of the circle.

So, ∠AOB = 2∠ACB

=> ∠ACB = ∠AOB/2

= 60^{0}/2

=> ∠ACB = 30^{0}

Also, ∠ADB = reflex ∠AOB/2

= (360^{0} – 60^{0})/2

= 300^{0}/2

= 150^{0}

**Question 3:**

In Fig. 10.37, ∠ PQR = 100^{0}, where P, Q and R are points on a circle with centre O. Find ∠ OPR

Answer:

Reflex angle POR = 2∠PQR

= 2 * 100^{0}

= 200^{0}

Now, angle POR = 360^{0} – 200^{0} = 160^{0}

Also, PO = OR [Radii of a circle]

∠OPR = ∠ORP [Opposite angles of isosceles triangle]

In ∆OPR,

∠POR = 160^{0}

So, ∠OPR = ∠ORP = 10^{0} [Angle sum property of a triangle]

**Question 4:**

In Fig. 10.38, ∠ ABC = 69^{0}, ∠ ACB = 31^{0}, find ∠ BDC.

Answer:

In Δ ABC, we have

∠ ABC + ∠ ACB + ∠ BAC = 180^{0} [Angle sum property of a triangle]

=> 69^{0} + 31^{0} + ∠ BAC = 180^{0}

=> 100^{0} + ∠ BAC = 180^{0}

=> ∠ BAC = 180^{0} – 100^{0}

=> ∠ BAC = 80^{0}

Also, ∠ BAC = ∠ BDC [Angles in the same segment]

So, ∠ BDC = 80^{0}

**Question 5:**

In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130^{0} and ∠ ECD = 20^{0}. Find ∠ BAC

Answer:

From the figure,

∠ BEC + ∠ DEC = 180^{0} [Linear pair]

=> 130^{0} + ∠ DEC = 180^{0}

=> ∠ DEC = 180^{0} - 130^{0}

=> ∠ DEC = 50^{0}

Now, in Δ DEC,

∠ DEC + ∠ DCE + ∠ CDE = 180^{0} [Angle sum property of a triangle]

=> 50^{0} + 20^{0} + ∠ CDE = 180^{0}

=> 70^{0} + ∠ CDE = 180^{0}

=> ∠ CDE = 180^{0} – 70^{0}

=> ∠ CDE = 110^{0}

Also, ∠ CDE = ∠ BAC [Angles in same segment]

=> ∠ BAC = 110^{0}

**Question 6:**

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70^{0}, ∠ BAC is 30^{0}, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

Answer:

From the figure,

∠CAD = ∠DBC= 70^{0} [Angles in the same segment]

Therefore, ∠DAB = ∠CAD + ∠BAC

= 70^{0} + 30^{0}

= 100^{0}

But, ∠DAB + ∠BCD = 180^{0} [Opposite angles of a cyclic quadrilateral]

So, ∠BCD = 180^{0} – 100^{0}

= 80^{0}

Now, we have AB = BC

Therefore, ∠BCA = 30^{0} [Opposite angles of an isosceles triangle]

Again, ∠DAB + ∠BCD = 180^{0} [Opposite angles of a cyclic quadrilateral]

=> 100^{0} + ∠BCA + ∠ECD = 180^{0} [Since ∠BCD = ∠BCA + ∠ECD]

=> 100^{0} + 30^{0} + ∠ECD = 180^{0}

=> 130^{0} + ∠ECD = 180^{0}

=> ∠ECD = 180^{0} – 130^{0}

=> ∠ECD = 50^{0}

Hence, ∠BCD = 80^{0} and ∠ECD = 50^{0}

**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.

Answer:

Given: ABCD is a cyclic quadrilateral, whose diagonals AC and BD are diameter of the circle

Passing through A, B, C and D.

To Prove: ABCD is a rectangle.

Proof:

In ∆AOD and ∆COB,

AO = CO [Radii of a circle]

OD = OB [Radii of a circle]

∠AOD = ∠COB [Vertically opposite angles]

So, ∆AOD ≅ ∆COB [by SAS axiom]

Hence, ∠OAD = ∠OCB [by CPCT]

But these are alternate interior angles made by the transversal AC, intersecting AD and BC.

So, AD || BC

Similarly, AB || CD.

Hence, quadrilateral ABCD is a parallelogram.

Also, ∠ABC = ∠ADC ………….(i) [Opposite angles of a parallelogram are equal]

And, ∠ABC + ∠ADC = 180^{0} ...(ii) [Sum of opposite angles of a cyclic quadrilateral is 180^{0}]

=> ∠ABC = ∠ADC = 90^{0} [From equation (i) and (ii)]

So, ABCD is a rectangle. [A parallelogram one of whose angles is 90^{0} is a rectangle]

**Question 8:**

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

Answer:

Given: A trapezium ABCD in which AB || CD and AD = BC.

To Prove: ABCD is a cyclic trapezium.

Construction: Draw DE ⊥ AB and CF ⊥ AB.

Proof:

In ∆DEA and ∆CFB, we have

AD = BC [Given]

∠DEA = ∠CFB = 90^{0} [DE ⊥ AB and CF ⊥ AB]

DE = CF [Distance between parallel lines remains constant]

So, ∆DEA ≅ ∆CFB [by RHS axiom]

=> ∠A = ∠B .........(i) [by CPCT]

and, ∠ADE = ∠BCF ……..(ii) [by CPCT]

Since, ∠ADE = ∠BCF [From equation (ii)]

=> ∠ADE + 90^{0} = ∠BCF + 90^{0 }=> ∠ADE + ∠CDE = ∠BCF + ∠DCF

=> ∠D = ∠C ………..(iii) [∠ADE + ∠CDE = ∠D, ∠BCF + ∠DCF = ∠C]

So, ∠A = ∠B and ∠C = ∠D ……(iv) [From (i) and (iii)]

=> ∠A + ∠B + ∠C + ∠D = 360^{0} [Sum of the angles of a quadrilateral is 360°]

=> 2(∠B + ∠D) = 360^{0} [Using equation (iv)]

=> ∠B + ∠D = 180^{0}

=> Sum of a pair of opposite angles of quadrilateral ABCD is 180^{0}.

=> ABCD is a cyclic trapezium

**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 and P,

Q respectively (see Fig. 10.40). Prove that ∠ ACP = ∠ QCD

Answer:

Given: 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.

To Prove: ∠ACP = ∠QCD.

Proof:

∠ACP = ∠ABP ..........(i) [Angles in the same segment]

∠QCD = ∠QBD ………(ii) [Angles in the same segment]

But, ∠ABP = ∠QBD ..(iii) [Vertically opposite angles]

From equation (i), (ii) and (ii), we get

∠ACP = ∠QCD

**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.

Answer:

Given: Sides AB and AC of a triangle ABC are diameters of two circles which intersect at D.

To Prove: D lies on BC.

Proof: Join AD

∠ADB = 90^{0} ..........(i) [Angle in a semicircle]

Also, ∠ADC = 90^{0} ………..(ii)

Adding equation (i) and (ii), we get

∠ADB + ∠ADC = 90^{0} + 90^{0}

=> ∠ADB + ∠ADC = 180^{0}

=> BDC is a straight line.

So, D lies on BC

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

**Question 11:**

ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Answer:

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To Prove: ∠CAD = ∠CBD

Proof: Let O be the mid-point of AC.

Then OA = OB = OC = OD

Mid point of the hypotenuse of a right triangle is equidistant from its vertices with O as centre

and radius equal to OA, draw a circle to pass through A, B, C and D.

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

Since, ∠CAD and ∠CBD are angles of the same segment.

Therefore, ∠CAD = ∠CBD.

**Question 12:**

Prove that a cyclic parallelogram is a rectangle.

Answer:

Given: ABCD is a cyclic parallelogram.

To prove: ABCD is a rectangle.

Proof: ∠ABC = ∠ADC ............(i) [Opposite angles of a parallelogram are equal]

But, ∠ABC + ∠ADC = 180^{0} ... (ii) [Sum of opposite angles of a cyclic quadrilateral is 180^{0}]

⇒ ∠ABC = ∠ADC = 90^{0} [From equation (i) and (ii)]

So, ABCD is a rectangle [A parallelogram one of whose angles is 90^{0} is a rectangle]

Hence, a cyclic parallelogram is a rectangle.

**Exercise 10.6 (Optional)**

**Question 1:**

Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Answer:

Given : Two intersecting circles, in which OO′ is the line of centres and A and B are two points of intersection.

To prove: ∠OAO′ = ∠OBO′

Construction: Join AO, BO, AO′ and BO′.

Proof: In ∆AOO′ and ∆BOO′, we have

AO = BO [Radii of the same circle]

AO′ = BO′ [Radii of the same circle]

OO′ = OO′ [Common]

So, ∆AOO′ ≅ ∆BOO′ [by SSS axiom]

=> ∠OAO′ = ∠OBO′ [by CPCT]

Hence, the line of centres of two intersecting circles subtends equal angles at the two points of

intersection.

**Question 2:**

Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre.

If the distance between AB and CD is 6 cm, find the radius of the circle.

Answer:

Let O be the centre of the circle and let its radius be r cm.

Draw OM ⊥ AB and OL ⊥ CD.

Then, AM = AB/2 = 5/2 cm

and, CL = CD/2 = 11/2 cm

Since, AB || CD, it follows that the points O, L, M are collinear and therefore, LM = 6 cm.

Let OL = x cm. Then OM = (6 – x) cm

Join OA and OC. Then OA = OC = r cm.

Now, from right-angled ∆OMA and ∆OLC, we have

OA^{2} = OM^{2} + AM^{2} and OC^{2} = OL^{2} + CL^{2} [By Pythagoras Theorem]

=> r^{2} = (6 – x)^{ 2} + (5/2)^{ 2} ……...(i)

and r^{2} = x^{2} + (11/2)^{ 2} ………... (ii)

From equation 1 and 2, we get

=> (6 – x)^{ 2} + (5/2)^{ 2} = x^{2} + (11/2)^{ 2}

=> 36 + x^{2} – 12x + 25/4 = x^{2} + 121/4

=> -12x = 121/4 – 25/4 – 36

=> -12x = 96/4 – 36

=> -12x = 24 – 36

=> 12x = – 12

=> x = 1

Substituting x = 1 in equation (i), we get

r^{2} = (6 – x)^{ 2} + (5/2)^{ 2}

=> r^{2} = (6 – 1)^{ 2} + (5/2)^{ 2}

=> r^{2} = 5^{2} + 25/4

=> r^{2} = 25 + 25/4

=> r^{2} = 125/4

=> r = √(125/4)

=> r = 5√5/2

Hence, radius r = 5√5/2 cm.

**Question 3:**

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?

Answer:

Let PQ and RS be two parallel chords of a circle with centre O.

We have, PQ = 8 cm and RS = 6 cm.

Draw perpendicular bisector OL of RS which meets PQ in M.

Since PQ || RS, therefore, OM is also perpendicular bisector of PQ.

Also, OL = 4 cm and RL = RS/2

=> RL = 3 cm

and PM = PQ/2

=> PM = 4 cm

In ∆ORL, we have

OR^{2} = RL^{2} + OL^{2} [Pythagoras theorem]

=> OR^{2} = 3^{2} + 4^{2} = 9 + 16

=> OR^{2} = 25

=> OR = 5 cm

Since OR = OP [Radii of the circle]

=> OP = 5 cm

Now, in ∆OPM

OM^{2} = OP^{2} – PM^{2} [Pythagoras theorem]

=> OM^{2} = 5^{2} – 4^{2} = 25 – 16 = 9

=> OM = √9 = 3 cm

Hence, the distance of the other chord from the centre is 3 cm.

**Question 4:**

Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle.

Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer:

Given: Two equal chords AD and CE of a circle with centre O. When meet at B when produced.

To Prove: ∠ABC = (∠AOC – ∠DOE)/2

Proof: Let ∠AOC = x, ∠DOE = y, ∠AOD = z

∠EOC = z [Equal chords subtends equal angles at the centre]

=> x + y + 2z = 36^{0} [Angle at a point] ………..(i)

OA = OD

=> ∠OAD = ∠ODA

Now in DOAD, we have

∠OAD + ∠ODA + z = 180^{0}

=> 2∠OAD = 180^{0} – z [Since ∠OAD = ∠OBA]

=> ∠OAD = 90^{0} – z/2 …………... (ii)

Similarly ∠OCE = 90° – z/2 …..... (iii)

=> ∠ODB = ∠OAD + ∠ODA [Exterior angle property]

=> ∠OEB = 90^{0} – z/2 + z [From equation (ii)]

=> ∠ODB = 90^{0} + z/2 ……….... (iv)

Also, ∠OEB = ∠OCE + ∠COE [Exterior angle property]

=> ∠OEB = 90^{0} – z/2 + z [From equation (iii)]

=> ∠OEB = 90^{0} + z/2 …….... (v)

Also, ∠OED = ∠ODE = 90^{0} – y/2 ……... (vi)

From equation (iv), (v) and (vi), we have

∠BDE = ∠BED = 90^{0} + z/2 – (90^{0} – y/2)

=> ∠BDE = ∠BED = 90^{0} + z/2 – 90^{0} + y/2

=> ∠BDE = ∠BED = (y + z)/2

=> ∠BDE = ∠BED = y + z ........ (vii)

So, ∠BDE = 180^{0} – (y + z)

=> ∠ABC = 180^{0} – (y + z) ......... (viii)

Now, (y - z)/2 = (360^{0} – y – 2z – y)/2 = 180^{0} – (y + z) …..... (ix)

From equation (viii) and (ix), we get

∠ABC = (x – y)/2

**Question 5:**

Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Answer:

Given: A rhombus ABCD whose diagonals intersect each other at O.

To prove: A circle with AB as diameter passes through O.

Proof: ∠AOB = 90^{0} [Diagonals of a rhombus bisect each other at 90^{0}]

=> ∆AOB is a right triangle right angled at O. => AB is the hypotenuse of A B right ∆ AOB.

=> If we draw a circle with AB as diameter, then it will pass through O. because angle is a

Semicircle is 90^{0} and ∠AOB = 90^{0}

**Question 6:**

ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Answer:

Given: ABCD is a parallelogram.

To Prove: AE = AD.

Construction: Draw a circle which passes through ABC and intersect CD (or CD produced) at E.

Proof: For figure (i), we get

∠AED + ∠ABC = 180^{0} [Linear pair] ......... (i)

But ∠ACD = ∠ADC = ∠ABC + ∠ADE

⇒ ∠ABC + ∠ADE = 180^{0} [From figure (ii)] .......... (ii)

From equation (i) and (ii), we get

∠AED + ∠ABC = ∠ABC + ∠ADE

=> ∠AED = ∠ADE

=> ∠AD = ∠AE [Sides opposite to equal angles are equal]

Similarly we can prove for Fig (ii) also.

**Question 7:**

AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.

Answer:

Given: A circle with chords AB and CD which bisect each other at O.

To Prove: (i) AC and BD are diameters

(ii) ABCD is a rectangle.

Proof: In ∆OAB and ∆OCD, we have

OA = OC [Given]

OB = OD [Given]

∠AOB = ∠COD [Vertically opposite angles]

=> ∆AOB ≅ ∠COD [SAS congruence]

=> ∠ABO = ∠CDO and ∠BAO = ∠BCO [CPCT]

=> AB || DC ........... (i)

Similarly, we can prove BC || AD ........... (ii)

Hence, ABCD is a parallelogram.

But ABCD is a cyclic parallelogram.

So, ABCD is a rectangle. [Proved in Q. 12 of Ex. 10.5]

=> ∠ABC = 90^{0} and ∠BCD = 90^{0}

=> AC is a diameter and BD is a diameter [Angle in a semicircle is 90^{0}]

**Question 8:**

Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90^{0} – A/2, 90^{0} – B/2 and 90^{0} – C/2.

Answer:

Given : ∆ABC and its circumcircle. AD, BE,

CF are bisectors of ∠A, ∠B, ∠C respectively.

Construction : Join DE, EF and FD.

Proof : We know that angles in the same

segment are equal.

So, ∠5 = ∠C/2 and ∠6 = ∠B/2 ………..(i)

∠1 = ∠A/2 and ∠2 = ∠C/2 ……….(ii)

∠4 = ∠A/2 and ∠3 = ∠B/2 ……..(iii)

From equation (i), we have

∠5 + ∠6 = ∠C/2 + ∠B/2

=> ∠D = ∠C/2 + ∠B/2 ............(iv) [Since ∠5 + ∠6 = ∠D]

But ∠A + ∠B + ∠C = 180^{0}

=> ∠B + ∠C = 180^{0} – ∠A

=> ∠B/2 + ∠C/2 = 90^{0} – ∠A/2

So, equation (iv) becomes,

∠D = 90^{0} – ∠A/2

Similarly, from equation (ii) and (iii), we can prove that

∠E = 90^{0} – ∠B/2 and ∠F = 90^{0} – ∠C/2

**Question 9:**

Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Answer:

Given: Two congruent circles which intersect at A and B. PAB is a line through A.

To Prove: BP = BQ.

Construction: Join AB.

Proof: AB is a common chord of both the circles.

But the circles are congruent.

=> arc ADB = arc AEB

=> ∠APB = ∠AQB [Angles subtended]

=> BP = BQ [Sides opposite to equal angles are equal]

**Question 10:**

In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answer:

Let angle bisector of ∠A intersect circumcircle of ∆ABC at D.

Now, join DC and DB.

∠BCD = ∠BAD [Angles in the same segment]

=> ∠BCD = ∠BAD = ∠A/2 [AD is bisector of ∠A] ….....(i)

Similarly ∠DBC = ∠DAC = ∠A/2 ………..... (ii)

From equation (i) and (ii), we get

∠DBC = ∠BCD

=> BD = DC [sides opposite to equal angles are equal]

=> D lies on the perpendicular bisector of BC.

Hence, angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of

∆ ABC.

.