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Since PQ is tangent to the circle at R, PO bisects the angles at ∠ EOR. Thus, EOP = POR = θ 1 (let) QO bisects the angle at FOR. Thus, FOQ = QOR = θ ₂ (let) Since AB and CD are parallel tangents to the circle, EF must be a straight line. Thus, from the diagram, ∠ EOP + POR + FOQ + QOR = 180° 2 ( θ 1 + θ 2 ) = 180° ( θ 1 + θ 2 ) = 90° ∠ EOP = 90°
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