NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

Table of symbols used in this chapter.

Symbol	Known As	Used for	Description
⊥	Uptag	Perpendicular to	A small horizontal line with a vertical line standing on it
′	Prime	Making two variables different like o & O′	Like inverted comma
≡	Identical To	Congruent to	3 horizontal lines as in =.
▶	Black upright Triangle	Triangle	A small upright triangle
∠	Angle	Angle	An angle

Dear students! this chapter contains 47 figures. Do not worry, each and every figure has been fully described by Professor T K Bansal, in such a fashion that you can easily visualize each figure without any difficulty. Have Courage, and proceed ahead with confidence.

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10.1 Introduction

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

We come across many objects in our daily life, which are round in shape, such as, wheels of a vehicle, bangles, dials of many clocks, coins of denominations 50 p, ₹ 1 and ₹ 5, key rings, buttons of shirts, etc. (see Figure 10.1).

Figure 10.1

 

In a clock, we observed that the second’s hand goes round the dial of the clock rapidly and its tip moves in a round path. This path traced by the tip of the second’s hand is called a circle. In this chapter, we will study about circles, other related terms, and some properties of a circle.

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10.2 Circles and Its Related Terms: A Review

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

Take a compass and fix a pencil in it. Put its pointed leg on a point on a sheet of a paper. Open the other leg to some distance. Keeping the pointed leg at the same point, rotate the other leg through one revolution. What is the closed figure traced by the pencil on paper? As we know, it is a circle (see Figure 10.2).

Figure 10.2

 

How did we get a circle? We kept one point fixed (at Point A in Figure 10.2) and drew all the points that were at a fixed distance from Point A. This gives us the following definition:

A Circle

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called a circle.

Centre and Radius of a circle

The fixed point is called the centre of the circle and the fixed distance is called the radius of the circle.

In Figure 10.3, O is the centre and the length OP is the radius of the circle.

Figure 10.3

 

We are already familiar with some of the following concepts from Class 6. We are just recalling them.

A circle divides the plane on which it lies into three parts. They are:
(i) inside the circle, which is also called the interior of the circle;
(ii) the circle itself and
(iii) outside the circle, which is also called the exterior of the circle (see Figure 10.4).

Figure 10.4

 

The circle and its interior make up the circular region.

Chord of a Circle

If we take two points P and Q on a circle, then the line segment PQ is called a chord of the circle (see Figure 10.5).

Diameter of a circle

The chord, which passes through the centre of the circle, is called a diameter of the circle. As in the case of radius, the word ‘diameter’ is also used in two senses, that is, as a line segment and also as its length.

Do you find any other chord of the circle that is longer than a diameter of the circle? No, we see that a diameter is the longest chord of the circle, and all diameters have the same length, which is equal to two times the radius of the circle. In Figure 10.5, AOB is a diameter of the circle.

How many diameters does a circle have? Draw a circle and see how many diameters you can find.

Fig. 10.5

 

An Arc of a Circle

A piece of a circle between two points is called an arc.

Look at the pieces of the circle between two points P and Q in Figure 10.6.

Figure 10.6

 

We find that there are two pieces, one longer and the other shorter (see Figure 10.7).

Figure 10.7

 

The longer one is called the major arc PQ and the shorter one is called the minor arc PQ. The minor arc PQ is also denoted by PQ arc ( where a small arc is drawn over PQ) and the major arc PQ by PRQ arc (where a a small arc is drawn over PRQ) , where R is some point on the arc between P and Q. Unless otherwise stated, arc PQ or arc PQ stands for minor arc PQ.

When P and Q are ends of a diameter, then both arcs are equal in length, and each is called a semicircle.

Circumference of a Circle

The length of the complete circle is called its circumference.

Segment of a Circle

The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle.

Here again, we find that there are two types of segments, which are the major segment and the minor segment (see Figure 10.8).

Fig 10.8

 

Sector of a Circle

The region between an arc and the two radii, joining the centre to the end points of the arc is called a sector.

Like segments, we find that the minor arc corresponds to the minor sector and the major arc corresponds to the major sector. In Figure 10.9, the region OPQ is the minor sector and remaining part of the circular region is the major sector.

When two arcs are equal, that is, each is a semicircle, then both segments and both sectors become the same and each is known as a semicircular region

Figure 10.9.

 

EXERCISE 10.1

Q1. Fill in the blanks:
(i) The centre of a circle lies in the ______ of the circle. (exterior/ interior)
(ii) A point, whose distance from the centre of a circle is greater than its radius lies in the ______ of the circle. (exterior/ interior)
(iii) The longest chord of a circle is a ______ of the circle. (radius/diameter)
(iv) An arc is a ______ when its ends are the ends of a diameter. (minor arc/semi circle)
(v) Segment of a circle is the region between an arc and ______ of the circle.(chord/diameter)
(vi) A circle divides the plane, on which it lies, in ______ parts. (1/2/3/4)

A1.
(i) Interior
(ii) Exterior
(iii) Diameter
(iv) Semicircle
(v) The chord
(vi) Three

Q2. 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.

A2.
(i) True
(ii) False, A circle has infinite number of equal chords.
(iii) False, If a circle is divided into three equal arcs, each one is a minor arc.
(iv) True (v) False, Sector is the region between the 2 radii and its corresponding arc.
(vi) True

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10.3 Angle Subtended by a Chord at a Point

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

Take a line segment PQ and a point R, not on the line containing PQ. Join PR and QR (see Figure 10.10).

Figure 10.10

 

Then ∠PRQ is called the angle subtended by the line segment PQ at the point R.

What are angles POQ, PRQ and PSQ called in Figure 10.11?

Figure 10.11

 

∠ POQ is the angle subtended by the chord PQ at the centre O, ∠ PRQ and ∠ PSQ are respectively the angles subtended by PQ at points R and S on the major and minor arcs PQ.

Let us examine the relationship between the size of the chord and the angle subtended by it at the centre. We can see by drawing different chords of a circle and angles subtended by them at the centre. It is observed that the longer is the chord, the bigger is the angle subtended by it at the centre.

What will happen if we take two equal chords of a circle? Will the angles subtended at the centre be the same or not?

Draw two or more equal chords of a circle and measure the angles subtended by them at the center (see Figure 10.12).

Figure 10.12

 

We find that the angles subtended by them at the centre are equal. Let us give a proof of this fact.

Theorem 10.1

Equal chords of a circle subtend equal angles at the centre.

Proof :

We are given two equal chords AB and CD of a circle with center O (see Figure 10.13). We want to prove that

\[∠AOB\ =\ ∠COD\]

Figure 10.13.

 

In triangles AOB and COD,
OA = OC (Radii of a circle)
OB = OD (Radii of a circle)
AB = CD (Given)

Therefore, ▶AOB ≡ ▶COD (SSS rule)

This gives ∠AOB = ∠COD (Corresponding parts of congruent triangles)

Remark :
For convenience, the abbreviation CPCT is used in place of ‘Corresponding parts of congruent triangles’, because we use this very frequently as we will see.

Now if two chords of a circle subtend equal angles at the centre, what can you say about the chords? Are they equal or not?

Let us examine this by the following activity:

Take a tracing paper and trace a circle on it. Cut it along the circle to get a disc. At its center O, draw an ∠AOB where A, B are points on the circle. Make another ∠POQ at the center = ∠AOB. Cut the disc along AB and PQ (see Figure 10.14).

Figure 10.14 (i)

 

Figure 10.14 (ii)

 

We get two segments ACB and PRQ of the circle. If we put one on the other, what do you observe? They completely cover each other, i.e., they are congruent. So, Chord AB = Chord PQ.

Though we have seen it for this particular case, we should try it out for many other equal angles too. The chords will all turn out to be equal because of the following theorem:

Theorem 10.2

If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

The above theorem is the converse of Theorem 10.1.

Note that in Figure 10.13, if we take ∠AOB = ∠COD, then

▶AOB ≡ ▶COD (Why?)

Can you now see that Chord AB = Chord CD?

EXERCISE 10.2

Q1. 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.

A1. Prove exactly as Theorem 10.1 by considering chords of congruent circles.

Q2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

A2. Use SAS axiom of congruence to show the congruence of the two triangles.

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10.4 Perpendicular from the centre to a chord

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

Activity :

Draw a circle on a tracing paper. Let O be its centre. Draw a chord AB. Fold the paper along a line through O so that one portion of the chord falls on the other. Let the crease cut AB at the point M. Then,

\[∠OMA\ =\ ∠OMB\ =\ 90°\]

and Segment OM is ⊥ Chord AB.

Does the point B coincide with Point A (see Figure 10.15)?

Figure 10.15

 

Yes, it does. So, M A = MB.

Give a proof yourself by joining OA and OB and proving the right triangles OMA and OMB to be congruent. This example is a particular instance of the following result:

Theorem 10.3

The perpendicular from the center of a circle to a chord bisects the chord.

What is the converse of this theorem? To write this, first let us be clear what is assumed in Theorem 10.3 and what is proved. Given that the perpendicular from the centre of a circle to a chord is drawn and to prove that it bisects the chord. Thus, in the converse, what the hypothesis is ‘if a line from the centre bisects a chord of a circle’ and what is to be proved is ‘ the line is perpendicular to the chord ’. So, the converse is:

Theorem 10.4

The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Is this true? Try it for few cases and see. You will see that it is true for these cases. See if it is true, in general, by doing the following exercise. We will write the stages and you give the reasons.

Let AB be a chord of a circle with center O and Point O is joined to the mid-point M of AB. We have to prove that OM ⊥ AB.

Join OA and OB (see Figure 10.16).

Figure 10.16.

 

In triangles OAM and OBM,
OA = OB (Why ?)
AM = BM (Why ?)
OM = OM (Common)
Therefore, ▶OAM ≡ ▶OBM (How ?)
This gives ∠OMA = ∠OMB = 90° (Why ?)

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10.5 Circle through Three Points

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

We have learnt in Chapter 6, that two points are sufficient to determine a line. That is, there is one and only one line passing through two points. A natural question arises. How many points are sufficient to determine a circle?

Take a point P. How many circles can be drawn through this point? We see that there may be as many circles as we like, passing through this point [see Figure 10.17(i)].

Fig. 10.17(i)

 

Now take two points P and Q. You again see that there may be an infinite number of circles passing through both the points P and Q simultaneously [see Figure 10.17(ii)].

Figure 10.17(i)

 

What will happen when we take three points A, B and C? Can we draw a circle passing through three collinear points?

No. If the points lie on a line, then the third point will lie either inside or outside the circle passing through two points (see Figure 10.18).

Figure 10.18

 

So, let us take three points A, B and C, which are not on the same line, or in other words, they are not collinear [see Figure 10.19(i)]. Draw perpendicular bisectors of AB and BC say, PQ and RS respectively. Let these perpendicular bisectors intersect at some point O (say). (Note that PQ and RS will intersect because they are not parallel) [see Figure 10.19(ii)].

Figure 10.19(i)

 

Figure 10.19(ii)

 

Now O lies on the perpendicular bisector PQ of chord AB, you have OA = OB, as every point on the perpendicular bisector of a line segment is equidistant from its end points, proved in Chapter 7.

Similarly, as O lies on the perpendicular bisector RS of chord BC, you get OB = OC

So OA = OB = OC, which means that the points A, B and C are at equal distances from the point O. So, if you draw a circle with centre O and radius OA, it will also pass through points B and C. This shows that there is a circle passing through the three points A, B and C. You know that two lines (perpendicular bisectors) can intersect at only one point, so you can draw only one circle with radius OA. In other words, there is a unique circle passing through A, B and C. You have now proved the following theorem:

Theorem 10.5

There is one and only one circle passing through three given non-collinear points.

Remark :
If ABC is a triangle, then by Theorem 10.5, there is a unique circle passing through the three vertices A, B and C of the triangle. This circle is called the circumcircle of the ▶ABC. Its centre and radius are called respectively the circumcentre and the circumradius of the triangle.

Example 1

Given an arc of a circle, complete the circle.

Solution :

Let arc PQ of a circle be given. We have to complete the circle, which means that we have to find its center and its radius.

Take a point R on the arc. Join PR and RQ. Use the construction that has been used in proving Theorem 10.5, to find the center and radius.

Taking the center and the radius so obtained, we can complete the circle (see Figure 10.20).

Figure 10.20

 

EXERCISE 10.3

Q1. Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

A1. 0, 1, 2. Two

Q2. Suppose you are given a circle. Give a construction to find its centre.

A2. Proceed as in Example 1.

Q3. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

A3. Join the centres O, O′ of the circles to the mid-point M of the common chord AB. Then, show ∠OMA = 90° and ∠O′MA = 90°.

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10.6 Equal Chords and Their Distances from the Centre

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

Let AB be a line and P be a point not collinear with AB. Since there are infinite number of points on a line, take a large number of points on line AB. If you join these points to P, you will get infinitely many line segments PL_1, PL_2, PM, PL_3, PL_4, etc. Which of these is the distance of Segment AB from Point P? You may think a while and get the answer. Out of these line segments, the perpendicular from P to AB, namely PM in Figure 10.21, will be the least. In Mathematics, we define this least length PM to be the distance of Line AB from Point P. So, you may say that:

Figure 10.21

 

The length of the perpendicular from a point to a line is the distance of the line from the point.

Note that if the point lies on the line, the distance of the line from the point is zero.

A circle can have infinitely many chords. You may observe by drawing chords of a circle that longer chord is nearer to the centre than the smaller chord. You may observe it by drawing several chords of a circle of different lengths and measuring their distances from the centre. What is the distance of the diameter, which is the longest chord from the centre? Since the centre lies on it, its disstance is zero. Do you think that there is some relationship between the length of chords and their distances from the centre? Let us see if this is so.

Activity :

Draw a circle of any radius on a tracing paper. Draw two equal chords AB and CD of it and also the perpendiculars OM and ON on them from the centre O. Fold the figure so that D falls on B and C falls on A [see Figure 10.22 (i)].

Figure 10.22 (i)

 

You may observe that Point O lies on the crease and Point N falls on Point M. Therefore, OM = ON. Repeat the activity by drawing congruent circles with centres O and O′ and taking equal chords AB and CD one on each. Draw perpendiculars OM and O′N on them [see Figure 10.22(ii)].

Figure 10.22 (ii)

 

Cut one circular disc and put it on the other so that AB coincides with CD. Then you will find that Point O coincides with Point O′, and Point M coincides with Point N. In this way you verified the following:

Theorem 10.6

Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Next, it will be seen whether the converse of this theorem is true or not. For this, draw a circle with centre O. From the centre O, draw two line segments O L and O M of equal length and lying inside the circle [see Figure 10.23(i)].

Figure 10.23(i)

 

Then draw chords PQ and RS of the circle perpendicular to OL and OM respectively [see Figure 10.23(ii)].

Figure 10.23(ii)

 

Measure the lengths of PQ and RS. Are these different? No, both are equal. Repeat the activity for more equal line segments and drawing the chords perpendicular to them. This verifies the converse of the Theorem 10.6 which is stated as follows:

Theorem 10.7

Chords equidistant from the center of a circle are equal in length.

We now take an example to illustrate the use of the above results:

Example 2

If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.

Solution :

Given that AB and CD are two chords of a circle, with center O, intersecting at a point E. PQ is a diameter through E, such that ∠AEQ = ∠DEQ (see Figure 10.24).

Figure 10.24

 

We have to prove that AB = CD. Draw perpendiculars OL and OM on chords AB and CD, respectively. Now

∠LOE = 180° − 90° − ∠LEO = 90° − ∠LEO (Angle sum property of a triangle)

= 90° − ∠AEQ = 90° − ∠DEQ

= 90° − ∠MEO = ∠MOE

In triangles OLE and OME,
∠LEO = ∠MEO (Why ?)
∠LOE = ∠MOE (Proved above)
EO = EO (Common)
Therefore, ▶OLE ≡ ▶OME (Why ?)
This gives OL = OM (CPCT)
So, AB = CD (Why ?)

EXERCISE 10.4

Q1. 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.

A1. 6 cm. First show that the line joining centres is perpendicular to the radius of the smaller circle and then that common chord is the diameter of the smaller circle.

Q2. 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.

A2. If AB and CD are equal chords of a circle with centre O intersecting at Point E, draw perpendiculars O M on AB and O N on CD and join OE. Show that right triangles OME and ONE are congruent.

Q3. 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.

A3. Proceed as in Example 2.

Q4. If a line intersects two concentric circles (circles with the same centre) with O as the centre, at points A, B, C and D, prove that AB = CD (see Figure 10.25).

Figure 10.25

 

A4. Draw perpendicular OM on AD.

Q5. 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?

A5. Represent Reshma, Salma and Mandip by R, S and M respectively. Let KR = x m (see figure). Area of triangle ORS = ½ x × 5. Also, area of triangle ORS = ½ RS × OL = ½ × 6 × 4. Find x and hence RM.

Q6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distances from each other on its boundary, each having a toy telephone in his hand to talk to each other. Find the length of the string of each phone.

A6. Use the properties of an equilateral triangle and also Pythagoras Theorem.

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10.7 Angle Subtended by an Arc of a Circle

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

We have seen that the end points of a chord other than diameter of a circle cuts it into two arcs : one major and other minor. If we take two equal chords, what can we say about the size of arcs? Is one arc made by first chord equal to the corresponding arc made by the other chord? In fact, they are more than just equal in length. They are congruent in the sense that if one arc is put on the other, without bending or twisting, one superimposes the other completely.

We can verify this fact by cutting the arc, corresponding to the chord CD from the circle along CD and put it on the corresponding arc made by equal chord AB. We will find that the arc CD superimposes the arc AB completely (see Figure 10.26).

Figure 10.26

 

This shows that equal chords make congruent arcs and conversely congruent arcs make equal chords of a circle. We can state it as follows:

If two chords of a circle are equal, then their corresponding arcs are congruent and conversely, if two arcs are congruent, then their corresponding chords are equal.

Also the angle subtended by an arc at the center is defined to be angle subtended by the corresponding chord at the center in the sense that the minor arc subtends the angle and the major arc subtends the reflex angle. Therefore, in Figure 10.27, the angle subtended by the minor arc PQ at O is ∠POQ and the angle subtended by the major arc PQ at O is reflex angle POQ.

Figure 10.27

 

In view of the property above and Theorem 10.1, the following result is true:

Congruent arcs (or equal arcs) of a circle subtend equal angles at the centre.

Therefore, the angle subtended by a chord of a circle at its centre is equal to the angle subtended by the corresponding (minor) arc at the centre. The following theorem gives the relationship between the angles subtended by an arc at the centre and at a point on the circle.

Theorem 10.8

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.

Proof :
Given an arc PQ of a circle subtending angles POQ at the center O, and PAQ at any point A on the remaining part of the circle. We need to prove that ∠POQ = 2 × ∠PAQ

Consider the three different cases as given in Figure 10.28. In (i), arc PQ is minor; in (ii), arc PQ is a semicircle and in (iii), arc PQ is major.

Figure 10.28 (i)

 

Figure 10.28 (ii)

 

Figure 10.28 (iii)

 

Let us begin by joining AO and extending it to a point B.

In all the cases, ∠BOQ = ∠OAQ + ∠AQO because an exterior angle of a triangle is equal to the sum of the two interior opposite angles. Also in ▶OAQ, OA = OQ (Radii of a circle) Therefore, ∠OAQ = ∠OQA (Theorem 7.5) This gives ∠BOQ = 2 × ∠OAQ (1) Similarly, ∠BOP = 2 × ∠OAP (2) From (1) and (2), ∠BOP + ∠BOQ = 2 × ( ∠OAP + ∠OAQ) This is the same as ∠POQ = 2 × ∠PAQ (3)

For the case (iii), where PQ is the major arc, (3) is replaced by reflex angle P O Q = 2 × ∠PAQ

Remark :
Suppose we join points P and Q and form a chord PQ in the above figures. Then ∠PAQ is also called the angle formed in the segment PAQ.

In Theorem 10.8, A can be any point on the remaining part of the circle. So if you take any other point C on the remaining part of the circle (see Figure 10.29),

Figure 10.29

 

We have ∠POQ = 2 × ∠PCQ = 2 × ∠PAQ Therefore, ∠PCQ = ∠PAQ.

This proves the following:

Theorem 10.9

Angles in the same segment of a circle are equal.

Again let us discuss the case (ii) of Theorem 10.8 separately.

Here ∠PAQ is an angle in the segment, which is a semicircle. Also, ∠PAQ = 1/2 ∠POQ = (1/ 2) × 180° = 90°.

If we take any other point C on the semicircle, again we get that ∠PCQ = 90° Therefore, we find another property of the circle as: Angle in a semicircle is a right angle. The converse of Theorem 10.9 is also true. It can be stated as:

Theorem 10.10

If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

We can see the truth of this result as follows: In Figure 10.30, AB is a line segment, which subtends equal angles at two points C and D. That is ∠ACB = ∠ADB

Figure 10.30

 

To show that the points A, B, C and D lie on a circle, let us draw a circle through the points A, C and B. Suppose it does not pass through the point D. Then it will intersect AD (or extended AD) at a point, say, E (or E′).

If points A, C, E and B lie on a circle,

∠ACB = ∠AEB (Why?) But it is given that ∠ACB = ∠ADB. Therefore, ∠AEB = ∠ADB. This is not possible unless E coincides with D. (Why?) Similarly, E′ should also coincide with D.

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10.8 Cyclic Quadrilaterals

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle (see Figure 10.31).

Figure 10.31

 

We find a peculiar property in such quadrilaterals. Draw several cyclic quadrilaterals of different sides and name each of these as ABCD. (This can be done by drawing several circles of different radii and taking four points on each of them.) Measure the opposite angles and write your observations in the following table.

Serial Number	∠A	∠B	∠C	∠D	∠A + ∠C	∠B + ∠D
1.
2.
3.
4.
5.
6.

What do we infer from the table?

We find that ∠A + ∠C = 180° and ∠B + ∠D = 180°, (neglecting the error in measurements). This verifies the following:

Theorem 10.11

The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.

In fact, the converse of this theorem, which is stated below is also true.

Theorem 10.12

If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.

We can see the truth of this theorem by following a method similar to the method adopted for Theorem 10.10.

Example 3

In Figure 10.32, AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E.

Figure 10.32

 

Prove that
∠AEB = 60°.

Solution :

Join OC, OD and BC.

▶ODC is equilateral (Why?)
Therefore, ∠COD = 60°
Now, ∠CBD = 1/2 ∠COD (Theorem 10.8)

This gives ∠CBD = 30° Again, ∠ACB = 90° (Why ?)
So, ∠BCE = 180° − ∠ACB = 90°
Which gives ∠CEB = 90° − 30° = 60°, i.e., ∠AEB = 60°

Example 4

In Figure 10.33, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠ DBC = 55° and ∠BAC = 45°, find ∠BCD.

Figure 10.33

 

Solution :

∠CAD = ∠DBC = 55° (Angles in the same segment)
Therefore, ∠DAB = ∠CAD + ∠BAC
= 55° + 45° = 100°

But ∠DAB + ∠BCD = 180° (Opposite angles of a cyclic quadrilateral)
So, ∠BCD = 180° − 100° = 80°

Example 5

Two circles intersect at two points A and B. AD and AC are diameters to the two circles (see Figure 10.34). Prove that B lies on the line segment DC.

Figure 10.34

 

Solution :
Join AB.
∠ABD = 90° (Angle in a semicircle)
∠ABC = 90° (Angle in a semicircle)
So, ∠ABD + ∠ABC = 90° + 90° = 180°
Therefore, DBC is a straight line. That is B lies on the line segment DC.

Example 6

Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic.”>

Solution :

Let, ABCD is a quadrilateral in which the angle bisectors AH, BF, CF and DH of internal angles A, B, C and D respectively form a quadrilateral EFGH. (See Figure 10.35)

Figure 10.35

 

Now, ∠FEH = ∠AEB = 180° − ∠EAB − ∠EBA (Why ?)
= 180° − 1/2( ∠A + ∠B)
and ∠FGH = ∠CGD = 180° − ∠GCD − ∠GDC (Why ?)
= 180° − ½ ( ∠ C + ∠ D)
Hence, ∠FEH + ∠ FGH = 180° − ½ ( ∠A + ∠B) + 180° − ½ ( ∠C + ∠D)
= 360° − ½ ( ∠A + ∠ B + ∠C + ∠D)
= 360° − ½ × 360°
= 360° − 180° = 180°
Therefore, by Theorem 10.12, the quadrilateral EFGH is cyclic.

EXERCISE 10.5

Q1. In Figure 10.36, A,B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Figure 10.36

 

A1. 45°

Q2. 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.

A2. 150°, 30°

Q3. In Figure 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Figure 10.37

 

A3. 10°

Q4. In Figure 10.38, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Figure 10.38

 

A4. 80°

Q5. In Figure 10.39, A, B, C and D are four points on a circle. AC and BD intersect at point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

Figure 10.39

 

A5. 110°

Q6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.

A6. ∠BCD = 80° and ∠ECD = 50°

Q7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Q8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

A8. Draw perpendiculars AM and BN on CD (AB ∥ CD and AB < CD). Show ▶AMD ≡ ▶BNC. This gives ∠C = ∠D and, therefore, ∠A + ∠C = 180°.

Q9. 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 Figure 10.40). Prove that ∠ACP = ∠QCD.

Figure 10.40

 

Q10. 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.

Q11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.

Q12. Prove that a cyclic parallelogram is a rectangle.

EXERCISE 10.6 [These exercises are not from examination point of view.]

Q1. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Q2. 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.

A2. Let O be the centre of the circle. Then perpendicular bisector of both the chords will be same and passes through O. Let r be the radius, then r^2 = (11/2)^2 + x^2 = (5/2)^2 + (6 − x)^2, where x is length of the perpendicular from O on the chord of length 11 cm. This gives x = 1. So, r = 5 square root 5/2 cm.

Q3. 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?

A3. 3 cm.

Q4. Let the vertex of an ∠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.

A4. Let ∠AOC = x and ∠DOE = y. Let ∠AOD = z. Then ∠EOC = z and x + y + 2z = 360°. ∠ODB = ∠OAD + ∠DOA = 90° − ½ z + z = 90° + ½ z. Also ∠OEB = 90° + ½z

Q5. Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Q6. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.

Q7. 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.

Q8. 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° - ½ A, 90° − 1/2B and 90° − 1/2C.

A8. ∠ABE = ∠ADE, ∠ADF = ∠ACF = ½ ∠C. Therefore, ∠EDF = ∠ABE + ∠ADF = ½ (∠B + ∠C) = ½ (180° − ∠A) = 90° − ½ ∠A.

Q9. 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.

A9. Use Q. 1, Ex. 10.2 and Theorem 10.8. 10. Let angle-bisector of ∠A intersect circumcircle of triangle ABC at D. Join DC and DB. Then ∠BCD = ∠BAD = ½ ∠A and ∠DBC = ∠DAC = ½ ∠A. Therefore, ∠BCD = ∠DBC, or DB = DC. So, D lies on the perpendicular bisector of BC.

Q10. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.

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10.9 Summary

NCERT Class 9 Mathematics for Blind Students made Screen Readable by Professor T K Bansal.

In this chapter, we studied following points:

1.
A circle is the collection of all points in a plane, which are equidistant from a fixed point in the plane.

2.
Equal chords of a circle (or of congruent circles) subtend equal angles at the centre.

3.
If the angles subtended by two chords of a circle (or of congruent circles) at the centre (corresponding centres) are equal, the chords are equal.

4.
The perpendicular from the center of a circle to a chord bisects the chord.

5.
The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

6.
There is one and only one circle passing through three non-collinear points.

7.
Equal chords of a circle (or of congruent circles) are equidistant from the centre (or corresponding centres).

8.
Chords equidistant from the center (or corresponding centers) of a circle (or of congruent circles) are equal.

9.
If two arcs of a circle are congruent, then their corresponding chords are equal and conversely if two chords of a circle are equal, then their corresponding arcs (minor, major) are congruent.

10.
Congruent arcs of a circle subtend equal angles at the centre.

11.
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.

12.
Angles in the same segment of a circle are equal.

13.
Angle in a semicircle is a right angle.

14.
If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.

15.
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°

16.
If sum of a pair of opposite angles of a quadrilateral is 180° the quadrilateral is cyclic.

End of Chapter.