Given: A is the centre of the circle. Tangents through external point D touch the circle at the points P and Q.
To prove: seg DP ≅ seg DQ
Construction: Draw seg AP and seg AQ.

Proof:
In ∆PAD and ∆QAD,
seg PA ≅ [segQA] [Radii of the same circle]
seg AD ≅ seg AD [Common side]

∠APD = ∠AQD = 90° [Tangent theorem]
∴ ∆PAD = ∆QAD [By Hypotenuse side test]
∴ seg DP = seg DQ [c.s.c.t]