Prove that the intercept of a tangent between 2 parallel tangents to a circle subtends a right angle at the centre.

Given: A circle with centre 'O' has PT and QR as two tangents parallel to each other touching the circles at T and R. PQ is the intercept between 2 tangents touching the circle at S.
To prove : ∠ POQ = 90°
Construction: Join OS and OT
Proof: Consider D POS and D POT
As OS is the radius at the point of contact where the tangent intercept touches the circle.
∠ OSP = 90°,
Similarly ∠OTP = 90°
i.e. ∠OSP = ∠OTP
Hypotenuse OP is common
OS = OT (radii of the same circle)
\\ D POT @ D SOP (RHS)
\\ POT = SOP (cpct)
Similarly D SOQ, D ROQ are congruent 
\\ ∠SOQ = ∠ROQ 
As TOR is a straight line
∠POT + ∠SOP + ∠SOQ + ∠ROQ = 180°
But ∠POT = ∠SOP and ∠SOQ = ∠ ROQ (already proved)
∠SOP + ∠SOP + ∠SOQ + ∠SOQ = 180°
\\ 2 ( ∠SOP + ∠SOQ) - 180°
∠SOP + ∠SOQ == 90°.