1. In the figure, A, B and C are three points on a circle with centre O such that ∠ BOC = 30º and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC. We have a circle with centre O, such that ∠ AOB = 60º and ∠BOC = 30º ∵ ∠AOB + ∠BOC = ∠AOC ∴ ∠AOC = 60º + 30º = 90º Now, the arc ABC subtends ∠AOC = 90º at the centre and ∠ADC at a point D on the circle other than the arc ABC. ∴ ∠ADC = (1/2)[∠AOC] ⇒ ∠ADC = (1/2)(90º) = 45º 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. We have a circle having a chord AB equal to radius of the circle. ∴ AO = BO = AB ⇒ ΔAOB is an equilateral triangle. Since, each angle of an equilateral = 60º. ⇒ ∠AOB = 60º Since, the arc ACB makes reflex ∠AOB = 360º ∠ 60º = 300º at the centre of the circle and ∠ABC at a point on the minor arc of the circle. ∴ ∠ACB = (1/2) [reflex ∠AOB] = (1/2)[300º] = 150º Similarly, ∠ADB = (1/2)[∠AOB] = (1/2)x [60º] = 30° Thus, the angle subtended by the chord on the minor arc = 150º and on the major arc = 30º. 3. In the figure, ∠ PQR = 100º, where P, Q and R are points on a circle with centre O. Find ∠OPR. ∵ The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference. ∴ reflex ∠POR = 2∠PQR But ∠PQR = 100º ∴ reflex ∠POR = 2 x 100º = 200º Since, ∠POR + reflex ∠POR = 360º ∴ ∠POR + 200º = 360º ⇒ ∠POR = 360º ∠ 200º ⇒ ∠POR = 160º Since, OP = OR [Radii of the same circle] ∴ In ΔPOR,∠OPR = ∠ORP [Angles opposite to equal sides of a triangle are equal] Also, ∠OPR + ∠ORP + ∠POR = 180º [Sum of the angles of a triangle = 180°] ⇒ ∠OPR + ∠OPR + 160º = 180º [∵ ∠OPR = ∠ORP] ⇒ 2∠OPR = 180º ∠ 160º = 20º ⇒ ∠OPR = (200/2)= 10º 4. In the figure, ∠ ABC = 69º, ∠ ACB = 31º, find ∠ BDC. We have, in ΔABC, ∠ABC = 69º and ∠ACB = 31º But ∠ABC + ∠ACB + ∠BAC = 180º ∴ 69º + 31º + ∠BAC = 180º ⇒ ∠BAC = 180º ∠ 69º ∠ 31º = 80º Since, angles in the same segment are equal. ∴ ∠BDC = ∠BAC ⇒ ∠BDC = 80º 5. In the figure, 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 = 20º. Find ∠BAC. In D CDE, Exterior ∠BEC = {Sum of interior opposite angles} [∵ BD is a straight line.] 130º = ∠EDC + ∠ECD 130º = ∠EDC + 20º ⇒ ∠EDC = 130º ∠ 20º = 110º ⇒ ∠BDC = 110º Since, angles in the same segment are equal. ∴ ∠BAC = ∠BDC ⇒ ∠BAC = 110º 6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70º, ∠BAC is 30º, find ∠BCD. Further, if AB = BC, find ∠ECD. Solution: ∵ Angles in the same segment of a circle are equal. ∴ ∠BAC = ∠BDC ⇒ 30º = ∠BDC Also ∠DBC = 70º [Given] ∴ In ΔBCD, we have ∠BCD + ∠DBC + ∠CDB = 180º [∵ Sum of angles of a triangle is 180º] ⇒ ∠BCD + 70º + 30º = 180º ⇒ ∠BCD = 180º ∠ 70º ∠ 30º = 80º Now, in ΔABC, ∵ AB = BC [Given] ∴ ∠BCA = ∠BAC [Angles opposite to equal sides of a triangle are equal] ⇒ ∠BCA = 30º [∵ ∠BAC = 30º] Now, ∠BCA + ∠ECD = ∠BCD ⇒ 30º + ∠ECD = 80º ⇒ ∠ECD = 80º ∠ 30º = 50º 7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle. ∵ AC and BD are diameters. ∴ AC = BD [∵ All diameters of a circle are equal] …(1) Since a diameter divides a circle into equal parts. ∴ ∠BAD = 90º [∵ Angle formed in a semicircle is 90º] Similarly, ∠ABC = 90º ∠BCD = 90º and ∠CDA = 90º Now, in right ΔABC and right ΔBAD, AC = BD [From (1)] AB = AB [Common] ∴ ΔABC ≌ ΔBAD [RHS criterion] ⇒ BC = AD [c.p.c.t.] Similarly, AB = CD Thus, the cyclic quadrilateral ABCD is such that its opposite sides are equal and each of its angle is right angle. ∴ ABCD is a rectangle. 8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic. We have a trapezium ABCD such that AB || CD and AD = BC. Let us draw BE || AD such that ABED is a parallelogram. ∵ The opposite angles of a parallelogram are equal. ∴ ∠BAD = ∠BED …(1) and AD = BE [Opposite sides of a parallelogram] …(2) But AD = BC [Given] …(3) ∴ From (2) and (3), we have BE = BC ⇒ ∠BEC = ∠BCE [Angles opposite to equal sides of a triangle D are equal] …(4) Now, ∠BED + ∠BEC = 180º [Linear pairs] ⇒ ∠BAD + ∠BCE = 180º [Using (1) and (4)] i.e. A pair of opposite angles of quadrilateral ABCD is 180º. ∴ ABCD is cyclic. ⇒ The trapezium ABCD is a cyclic. 9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the figure). Prove that ∠ACP = ∠QCD. Since angles in the same segment of a circle are equal. ∴ ∠ACP = ∠ABP …(1) Similarly, ∠QCD = ∠QBD …(2) Since ∠ABP = ∠QBD [Vertically opposite angles are equal] ∴ From (1) and (2), we have ∠ACP = ∠QCD 10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side. We have a DABC, and two circles described with diameters as AB and AC respectively. They intersect in a point D. Let us join A and D. ∵ AB is a diameter. ∴ ∠ADB is an angle formed in a semicircle. ⇒ ∠ADB = 90º …(1) Similarly, ∠ADC = 90º …(2) Adding (1) and (2), we have ∠ADB + ∠ADC = 90º + 90º = 180º i.e. B, D and C are collinear points. ⇒ BC is a straight line. Thus, D lies on BC. 11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD. We have right ΔABC and right ΔADC such that they are having AC as their common hypotenuse. ∵ AC is a hypotenuse. ∵ ∠ADC = 90º = ∠ABC ∴ Both the triangles are in the same semicircle. ⇒ A, B, C and D are concyclic. Let us join B and D. ∵ DC is a chord. ∴ ∠CAD and ∠CBD are formed in the same segment. ⇒ ∠CAD = ∠CBD. 12. Prove that a cyclic parallelogram is a rectangle. We have a cyclic parallelogram ABCD. Since, ABCD is a cyclic quadrilateral. ∴ Sum of its opposite angles = 180º ⇒ ∠A + ∠C = 180º …(1) But ∠A = ∠C …(2) [∵ Opposite angles of parallelogram are equal] From (1) and (2), we have ∠A= ∠C = 90º Similarly, ∠B= ∠D = 90º ⇒ Each angle of the parallelogram ABCD is of 90º. Thus, ABCD is a rectangle.