∵ DE || AC, By basic proportionality theorem
Also, DC || AP
From  and , we get
Let TP and TQ are two tangents of a circle at points P and Q respectively with center O.
To prove: ∠PTQ = 2∠OPQ
Let ∠PTQ = θ
As lengths of tangents drawn from an external point to the circle are equal, therefore TP = TQ.
∴ ΔPQT is an isosceles triangle.
∴ ∠TPQ = ∠TQP = 1/2(180° - θ) = 90° – 1/2θ
Also, tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴ ∠OPQ = ∠OPT – ∠TPQ = 90° – (90°– 1/2θ)
Thus, ∠PTQ = 2∠OPQ.
Rate this question :