Tangent at P to the circumcircle of triangle PQR is drawn. If the tangent is parallel to side, QR show that ΔPQR is isosceles.
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.