Circles

Welcome dear students! Today we are going to learn about Circles from Class 9 Maths.

Let us begin with Section 9.1, Angle Subtended by a Chord at a Point. You have already studied about circles and its parts in Class 6. Take a line segment PQ and a point R not on the line containing PQ. Join PR and QR. Then ∠PRQ is called the angle subtended by the line segment PQ at the point R. In this figure, we see a line segment PQ with a point R above it, forming a triangle PQR. Now, look at Figure 9.2. What are ∠POQ, ∠PRQ and ∠PSQ called? ∠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. In this figure, we see a circle with centre O. A chord PQ divides the circle into two arcs. Point R is on the major arc and point S is on the minor arc. Lines are drawn from P and Q to O, R, and S, forming the respective angles.

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Let us examine the relationship between the size of the chord and the angle subtended by it at the centre. You may see by drawing different chords of a circle and angles subtended by them at the centre that the longer is the chord, the bigger will be the angle subtended by it at the centre. What will happen if you 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 centre. You will find that the angles subtended by them at the centre are equal. Let us give a proof of this fact. Theorem 9.1 states: Equal chords of a circle subtend equal angles at the centre. Proof: You are given two equal chords AB and CD of a circle with centre O. You want to prove that ∠AOB = ∠COD. In ∆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 (CPCT). For convenience, the abbreviation CPCT will be used in place of ‘Corresponding parts of congruent triangles’.

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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 centre O, draw an ∠AOB where A and B are points on the circle. Make another ∠POQ at the centre equal to ∠AOB. Cut the disc along AB and PQ. You will get two segments ACB and PRQ of the circle. If you put one on the other, what do you observe? They cover each other, that is, they are congruent. So AB = PQ. Though you have seen it for this particular case, try it out for other equal angles too. The chords will all turn out to be equal because of the following theorem. Theorem 9.2 states: If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal. This is the converse of Theorem 9.1. Note that in Figure 9.4, if you take ∠AOB = ∠COD, then ∆AOB ≅ ∆COD. Can you now see that AB = CD?

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Let us solve Exercise 9.1. 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. Solution: Let there be two congruent circles with centres O and O′. Let AB be a chord of the first circle and CD be a chord of the second circle such that AB = CD. We need to prove that ∠AOB = ∠CO′D. In ∆AOB and ∆CO′D, OA = O′C (Radii of congruent circles), OB = O′D (Radii of congruent circles), AB = CD (Given). Therefore, by SSS rule, ∆AOB ≅ ∆CO′D. 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. Solution: Let the two congruent circles have centres O and O′. Let chord AB subtend ∠AOB at centre O, and chord CD subtend ∠CO′D at centre O′. Given that ∠AOB = ∠CO′D. In ∆AOB and ∆CO′D, OA = O′C and OB = O′D as radii of congruent circles. The included angles ∠AOB and ∠CO′D are equal. Therefore, by SAS rule, ∆AOB ≅ ∆CO′D. Hence, AB = CD by CPCT.

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Next, we proceed to Section 9.2, Perpendicular from the Centre to a Chord. 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 a portion of the chord falls on the other. Let the crease cut AB at the point M. Then, ∠OMA = ∠OMB = 90° or OM is perpendicular to AB. Does the point B coincide with A? Yes it will. So MA = 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 9.3 states: The perpendicular from the centre 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 9.3 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, 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 9.4 states: 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 centre O and O is joined to the mid-point M of AB. You have to prove that OM ⊥ AB. Join OA and OB. In ∆OAM and ∆OBM, OA = OB (Radii of the same circle), AM = BM (M is the mid-point), OM = OM (Common). Therefore, ∆OAM ≅ ∆OBM (SSS rule). This gives ∠OMA = ∠OMB = 90° (CPCT and linear pair).

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Now we move to Section 9.3, Equal Chords and their Distances from the Centre. Let AB be a line and P be a point. Since there are infinite numbers of points on a line, if you join these points to P, you will get infinitely many line segments PL₁, PL₂, PM, PL₃, PL₄, etc. Which of these is the distance of AB from 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 9.8, will be the least. In Mathematics, we define this least length PM to be the distance of AB from P. So you may say that: 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, the distance 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.

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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. You may observe that O lies on the crease and N falls on 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. Cut one circular disc and put it on the other so that AB coincides with CD. Then you will find that O coincides with O′ and M coincides with N. In this way you verified the following: Theorem 9.5 states: 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 OL and OM of equal length and lying inside the circle. Then draw chords PQ and RS of the circle perpendicular to OL and OM respectively. 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 Theorem 9.5 which is stated as follows: Theorem 9.6 states: Chords equidistant from the centre of a circle are equal in length.

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We now take an example to illustrate the use of the above results. Example 1: 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 centre O intersecting at a point E. PQ is a diameter through E, such that ∠AEQ = ∠DEQ. You 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 ∆OLE and ∆OME, ∠LEO = ∠MEO (Given), ∠LOE = ∠MOE (Proved above), EO = EO (Common). Therefore, ∆OLE ≅ ∆OME (AAS rule). This gives OL = OM (CPCT). So, AB = CD (Chords equidistant from centre are equal).

Let us now solve Exercise 9.2. 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. Solution: Let the two circles have centres O and O′ with radii OA = 5 cm and O′A = 3 cm. The distance OO′ = 4 cm. Let the common chord be AB, intersecting OO′ at M. Since the line joining the centres is the perpendicular bisector of the common chord, OM ⊥ AB and AM = MB. Let OM = x. Then O′M = 4 – x. In right ∆OMA, OA² = OM² + AM², so 25 = x² + AM². In right ∆O′MA, O′A² = O′M² + AM², so 9 = (4 – x)² + AM². Subtracting the second equation from the first gives 16 = x² – (4 – x)². This simplifies to 16 = x² – (16 + x² – 8x), which gives 16 = 8x – 16, so 32 = 8x, giving x = 4. This means OM = 4 cm. Substituting x = 4 into the first equation: 25 = 16 + AM², so AM² = 9, giving AM = 3 cm. Therefore, the length of the common chord AB = 2 × AM = 6 cm.

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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. Solution: Let equal chords AB and CD intersect at point E inside the circle with centre O. Draw perpendiculars OM and ON from O to AB and CD respectively. Since AB = CD, OM = ON by Theorem 9.5. In right ∆OME and ∆ONE, OM = ON, OE = OE (Common), and ∠OME = ∠ONE = 90°. Therefore, ∆OME ≅ ∆ONE by RHS rule. Hence, ME = NE by CPCT. Since OM bisects AB and ON bisects CD, AM = MB and CN = ND. Thus, AE = AM – ME and CE = CN – NE. Since AM = CN and ME = NE, AE = CE. Similarly, EB = ED. Hence, the segments are equal. 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. Solution: Using the same figure and setup, we have already proved that ∆OME ≅ ∆ONE. Therefore, ∠OEM = ∠OEN by CPCT. Since OE is the line joining the intersection point E to the centre O, it makes equal angles with the chords AB and CD. Question 4: If a line intersects two concentric circles with centre O at A, B, C and D, prove that AB = CD. Solution: Let the line intersect the outer circle at A and D, and the inner circle at B and C. Draw a perpendicular OM from centre O to the line AD. Since OM ⊥ chord AD, it bisects AD, so AM = MD. Similarly, OM ⊥ chord BC, so it bisects BC, giving BM = MC. Now, AB = AM – BM, and CD = MD – MC. Since AM = MD and BM = MC, it follows that AB = CD.

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Question 5: Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m 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 6 m each, what is the distance between Reshma and Mandip? Solution: Let the circle have centre O and radius 5 m. Let Reshma be at R, Salma at S, and Mandip at M. Given RS = SM = 6 m. We need to find RM. Since RS = SM, chords RS and SM are equal, so they are equidistant from O. Let OS intersect RM at N. Since ∆RSM is isosceles with RS = SM, SN is perpendicular to RM and bisects it. In ∆ORS, OR = OS = 5 m, RS = 6 m. Let ON = x, then SN = 5 – x. In right ∆RNS, RN² + SN² = RS². In right ∆RNO, RN² + ON² = OR². Let RN = y. Then y² + (5 – x)² = 36, and y² + x² = 25. Subtracting gives 25 – 10x + x² – x² = 36 – 25, so 25 – 10x = 11, giving 10x = 14, so x = 1.4. Then y² = 25 – 1.96 = 23.04, so y = 4.8. Thus RM = 2y = 9.6 m. Question 6: A circular park of radius 20 m 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. Solution: Let the boys be at points A, S, and D on a circle of radius 20 m. Since they are at equal distances, ∆ASD is equilateral. Let the side length be a. The distance from the centre O to each vertex is 20 m. In an equilateral triangle, the circumradius R = a / √3. So 20 = a / √3. Therefore, a = 20√3 m. The length of each string is 20√3 m.

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Now we proceed to Section 9.4, Angle Subtended by an Arc of a Circle. You 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 you take two equal chords, what can you say about the size of arcs? Is one arc made by first chord equal to the corresponding arc made by another 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. You 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. You will find that the arc CD superimposes the arc AB completely. This shows that equal chords make congruent arcs and conversely congruent arcs make equal chords of a circle. You 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 centre is defined to be angle subtended by the corresponding chord at the centre in the sense that the minor arc subtends the angle and the major arc subtends the reflex angle. Therefore, in Figure 9.14, 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 ∠POQ. In view of the property above and Theorem 9.1, 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.

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The following theorem gives the relationship between the angles subtended by an arc at the centre and at a point on the circle. Theorem 9.7 states: 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 centre O and ∠PAQ at a 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 9.15. In case (i), arc PQ is minor; in case (ii), arc PQ is a semicircle and in case (iii), arc PQ is major. 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. Similarly, ∠BOP = 2∠OAP. From these two equations, ∠BOP + ∠BOQ = 2(∠OAP + ∠OAQ). This is the same as ∠POQ = 2∠PAQ. For case (iii), where PQ is the major arc, the equation is replaced by reflex ∠POQ = 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 PAQP. In Theorem 9.7, 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, you have ∠POQ = 2∠PCQ = 2∠PAQ. Therefore, ∠PCQ = ∠PAQ. This proves the following: Theorem 9.8 states: Angles in the same segment of a circle are equal.

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Again let us discuss case (ii) of Theorem 9.8 separately. Here ∠PAQ is an angle in the segment, which is a semicircle. Also, ∠PAQ = 1/2 ∠POQ = 1/2 × 180° = 90°. If you take any other point C on the semicircle, again you get that ∠PCQ = 90°. Therefore, you find another property of the circle as: Angle in a semicircle is a right angle. The converse of Theorem 9.8 is also true. It can be stated as: Theorem 9.9 states: 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). You can see the truth of this result as follows: In Figure 9.17, AB is a line segment, which subtends equal angles at two points C and D. That is ∠ACB = ∠ADB. 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. But it is given that ∠ACB = ∠ADB. Therefore, ∠AEB = ∠ADB. This is not possible unless E coincides with D. Similarly, E′ should also coincide with D.

Now we move to Section 9.5, Cyclic Quadrilaterals. A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle. You will find a peculiar property in such quadrilaterals. Draw several cyclic quadrilaterals of different sides and name each of these as ABCD. Measure the opposite angles and write your observations. What do you infer from the measurements? You find that ∠A + ∠C = 180° and ∠B + ∠D = 180°, neglecting the error in measurements. This verifies the following: Theorem 9.10 states: 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 9.11 states: If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic. You can see the truth of this theorem by following a method similar to the method adopted for Theorem 9.9.

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Let us now solve the worked examples. Example 2: In Figure 9.19, 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. Prove that ∠AEB = 60°. Solution: Join OC, OD and BC. ∆ODC is equilateral because OD = OC = CD, all being radii or equal to radius. Therefore, ∠COD = 60°. Now, ∠CBD = 1/2 ∠COD by Theorem 9.7. This gives ∠CBD = 30°. Again, ∠ACB = 90° because it is an angle in a semicircle. So, ∠BCE = 180° – ∠ACB = 90°. Which gives ∠CEB = 90° – 30° = 60°, that is, ∠AEB = 60°. Example 3: In Figure 9.20, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD. Solution: ∠CAD = ∠DBC = 55° because they are angles in the same segment. Therefore, ∠DAB = ∠CAD + ∠BAC = 55° + 45° = 100°. But ∠DAB + ∠BCD = 180° because they are opposite angles of a cyclic quadrilateral. So, ∠BCD = 180° – 100° = 80°. Example 4: Two circles intersect at two points A and B. AD and AC are diameters to the two circles. Prove that B lies on the line segment DC. Solution: Join AB. ∠ABD = 90° because it is an angle in a semicircle. ∠ABC = 90° because it is an 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.

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Example 5: Prove that the quadrilateral formed, if possible, by the internal angle bisectors of any quadrilateral is cyclic. Solution: In Figure 9.22, 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. Now, ∠FEH = ∠AEB = 180° – ∠EAB – ∠EBA = 180° – 1/2(∠A + ∠B). And ∠FGH = ∠CGD = 180° – ∠GCD – ∠GDC = 180° – 1/2(∠C + ∠D). Therefore, ∠FEH + ∠FGH = 180° – 1/2(∠A + ∠B) + 180° – 1/2(∠C + ∠D). This equals 360° – 1/2(∠A + ∠B + ∠C + ∠D). Since the sum of angles in a quadrilateral is 360°, this equals 360° – 1/2 × 360° = 360° – 180° = 180°. Therefore, by Theorem 9.11, the quadrilateral EFGH is cyclic.

Now let us solve Exercise 9.3. Question 1: In Figure 9.23, 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. Solution: ∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°. The angle subtended by arc AC at the centre is ∠AOC = 90°. By Theorem 9.7, the angle subtended by the same arc at any point on the remaining part of the circle is half of it. Therefore, ∠ADC = 1/2 × 90° = 45°. 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: Let the chord be AB and centre O. Given AB = OA = OB. So ∆OAB is equilateral. Hence, ∠AOB = 60°. The angle subtended at a point on the major arc is half of ∠AOB, which is 30°. The angle subtended at a point on the minor arc is half of the reflex ∠AOB. Reflex ∠AOB = 360° – 60° = 300°. So the angle on the minor arc = 1/2 × 300° = 150°. Question 3: In Figure 9.24, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR. Solution: The angle subtended by arc PR at the circumference is ∠PQR = 100°. The angle subtended at the centre is reflex ∠POR = 2 × 100° = 200°. So ∠POR = 360° – 200° = 160°. In ∆OPR, OP = OR as radii, so it is isosceles. Therefore, ∠OPR = ∠ORP. The sum of angles in ∆OPR is 180°, so 2∠OPR + 160° = 180°. This gives 2∠OPR = 20°, so ∠OPR = 10°.

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Question 4: In Figure 9.25, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC. Solution: In ∆ABC, ∠BAC = 180° – 69° – 31° = 80°. Angles ∠BAC and ∠BDC are subtended by the same arc BC in the same segment. Therefore, ∠BDC = ∠BAC = 80°. Question 5: In Figure 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC. Solution: In ∆ECD, ∠CED = 180° – ∠BEC because they form a linear pair, so ∠CED = 50°. In ∆ECD, ∠EDC = 180° – 50° – 20° = 110°. Angles ∠BAC and ∠BDC subtend the same arc BC. ∠BDC = ∠EDC = 110°. Since angles in the same segment are equal, ∠BAC = ∠BDC = 110°. Question 6: 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. Solution: ∠DAC = ∠DBC = 70° because they are in the same segment. So ∠DAB = ∠DAC + ∠BAC = 70° + 30° = 100°. In cyclic quadrilateral, ∠DAB + ∠BCD = 180°. So ∠BCD = 80°. If AB = BC, then ∆ABC is isosceles, so ∠BCA = ∠BAC = 30°. Now ∠ECD = ∠BCD – ∠BCA = 80° – 30° = 50°. 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: Let the cyclic quadrilateral be ABCD with diagonals AC and BD as diameters. Since AC is a diameter, ∠ABC and ∠ADC are angles in a semicircle, so they are 90° each. Since BD is a diameter, ∠BAD and ∠BCD are angles in a semicircle, so they are 90° each. Thus, all four angles are 90°. A quadrilateral with all angles 90° is a rectangle.

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Question 8: If the non-parallel sides of a trapezium are equal, prove that it is cyclic. Solution: Let ABCD be a trapezium with AB || DC and AD = BC. Draw CE || AD, meeting AB at E. Then AECD is a parallelogram, so AD = CE and ∠ADC = ∠AEC. Since AD = BC, CE = BC, so ∆CEB is isosceles, giving ∠CEB = ∠CBE. Now ∠AEC + ∠CEB = 180°. Substituting, ∠ADC + ∠ABC = 180°. Since one pair of opposite angles sums to 180°, ABCD is cyclic. 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. Prove that ∠ACP = ∠QCD. Solution: Join AC, CP, CQ, CD. In circle one, ∠ACP and ∠ABP subtend arc AP. In circle two, ∠QCD and ∠QBD subtend arc QD. Since ABD and PBQ are straight lines intersecting at B, ∠ABP = ∠DBQ as vertically opposite angles. Also, ∠ACP = ∠ABP because they are in the same segment of the first circle. And ∠QCD = ∠QBD because they are in the same segment of the second circle. Therefore, ∠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. Solution: Let ∆ABC be given. Draw circles with diameters AB and AC. Let them intersect at A and D. Join AD. Since AB is a diameter, ∠ADB = 90°. Since AC is a diameter, ∠ADC = 90°. Therefore, ∠ADB + ∠ADC = 180°. This means BDC is a straight line. Hence, D lies on BC, the third side. Question 11: ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD. Solution: Since ∠ABC = 90° and ∠ADC = 90°, points B and D lie on the circle with diameter AC. Thus, ABCD is a cyclic quadrilateral. Angles ∠CAD and ∠CBD are subtended by the same arc CD. Therefore, ∠CAD = ∠CBD. Question 12: Prove that a cyclic parallelogram is a rectangle. Solution: Let ABCD be a cyclic parallelogram. In a parallelogram, opposite angles are equal, so ∠A = ∠C. In a cyclic quadrilateral, opposite angles sum to 180°, so ∠A + ∠C = 180°. Substituting, 2∠A = 180°, so ∠A = 90°. A parallelogram with one right angle is a rectangle.

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Finally, let us review the summary of this chapter. In this chapter, you have studied the following points. One, a circle is the collection of all points in a plane, which are equidistant from a fixed point in the plane. Two, equal chords of a circle (or of congruent circles) subtend equal angles at the centre. Three, if the angles subtended by two chords of a circle (or of congruent circles) at the centre (or corresponding centres) are equal, the chords are equal. Four, the perpendicular from the centre of a circle to a chord bisects the chord. Five, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord. Six, equal chords of a circle (or of congruent circles) are equidistant from the centre (or corresponding centres). Seven, chords equidistant from the centre (or corresponding centres) of a circle (or of congruent circles) are equal. Eight, 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. Nine, congruent arcs of a circle subtend equal angles at the centre. Ten, 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. Eleven, angles in the same segment of a circle are equal. Twelve, angle in a semicircle is a right angle. Thirteen, 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. Fourteen, the sum of either pair of opposite angles of a cyclic quadrilateral is 180°. Fifteen, if sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]