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Question
Tangent at P to the circumcircle of triangle PQR is drawn. If the tangent is parallel to side, QR show that ΔPQR is isosceles.
Solution
DE is the tangent to the circle at P.
DE || QR ...(Given)
∠EPR = ∠PRQ ...(Alternate angles are equal)
∠DPQ = ∠PQR (Alternate angles are equal) ...(i)
Let ∠DPQ = x and ∠EPR = y
Since the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment
∴ ∠DPQ = ∠PRQ ...(ii) (DE is tangent and PQ is chord)
From (i) and (ii)
∠PQR = ∠PRQ
`=>` PQ = PR
Hence, triangle PQR is an isosceles triangle.
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