Geometry Angles In a Alternate Segment

 Angles In a Alternate SegmentThe angle between the tangent and chord at the point of contact is equal to the angle in the alternate segment.
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Example: In right ABC, a circle with side AB as diameter is drawn to intersect the hypotenuse AC at P. Prove that the tangent to the circle at P bisects the side BC.
Solution: Let O be the centre of the given circle.
 Suppose the tangent at P meets BC at Q. Join BP. You have to prove that BQ = QC. Since BPQ and BAP = BAC are angles in the alternate segments of chord BP. BPQ = BAC ...(i) Since AB is a diameter of the circle and angle in a semi-circle is right angle. APB = 90 BPC = 90 ...[APB and BPC are linear pairs] ...(ii) Since A ABC is a right triangle, right angled at B. BAC + BCA = 90 ...(iii) From equations (ii) and (iii), you get BAC + BCA = BPC BAC + BCA = BPQ + QPC ...[BPC = BPQ + QPC] BAC + BCA = BAC + QPC ...[BPQ = BAC] BCA = QPC QCA = QPC ...[BCA = QCA] Thus, in A PQC, you have QCA = QPC PQ = QC [ Sides opposite to equal angles are equal] ...(iv) Since tangents from an exterior point to a circle are equal in length. QP = QB i.e. PQ = QB ...(v) From equations (iv) and (v), you get BQ = QC. Thus, Q is a point on BC such that BQ = QC. Hence, PQ bisects BC.

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