In Fig. 2 DE || A

∵ DE || AC, By basic proportionality theorem

[1]

Also, DC || AP

[2]

From [1] and [2], we get

Hence, proved!

OR

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)

= 1/2θ

= 1/2∠PTQ

Thus, ∠PTQ = 2∠OPQ.

Hence, proved.