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Geometry Angles In a Alternate Segment
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Angles In a Alternate Segment
The 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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