We know that sum of opposite () angles of cyclic quadrilateral is 180° In the figure, ∠ADC = 130° and chord BC = chord BE, Find ∠CBE. Thus, angle subtended by chord at any point in major segment = 30° Let AB is chord of circle chord AB = radius OA radius OB () find the angle subtended by this chord in major segment at any point We know thatĪ chord and radius of a circle are equal. Solution : O is circumcentre of ∆ABC, join OB and OC. In O is circurncenter of ∆ABC, Prove () that ∠OBC + ∠BAC = 90° Which makes an angles 90° and 150° with centre Find ∠BAC if AB and AC lie on opposite sides of center.ĪB and AC arc two chords of circle. Two chords AB and AC subtend angle 90° and 150° respectively () at center of the circle. Thus, angles subtended by minor arc BC is same segment will be equal. Taking AB as diameter, draw a circle which passes through points C and B If two triangles ∆ACB and ∆ADB are drawn () on both sides of common hypotenuse AB, then Prove that ∠BAC = ∠BDC. ∵ Angle subtended by arc BC at the center of circle i.e., ∠BOC is double the angle subtended by same arc at point A of remaining part of circle i.e., ∠BAC. ∆ABC whose circumcenter is point O and D is mid point of BCĬonstruction: Draw OD ⊥ BC and join OB and OC If O is circumcenter of ∆ABC and D is mid point of base BC, then prove that ∠BOD = ∠A. We know that angle subtended by an arc of a () circle at center is double the angle subtended at remaining part of circle by same arc. In figure, if ∠ABC 45°, then prove that OA ⊥ OC.

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Ex 12.3 class 10