In Fig. 10.13, XY and X^'Y^' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X^'Y^' at B. Prove that ∠AOB = 90°.

Joining Points O and C we get the figure:

Now considering the triangles △OPA and △OCA,

(i) OP = OC (radius of the same circle)

(ii) AO = AO (common side)

(iii) AP = AC (tangents from external point A)

So, △OPA ≅ △OCA (using SSS congruency)

So, ∠POA = ∠COA … (i)

Similarly,

△OQB ≅ △OCB

Thus, ∠QOB = ∠COB … (ii)

Since the line POQ is a straight line, it can be considered as the diameter of the circle.

So, ∠POA + ∠COA + ∠COB +∠QOB = 180°

Now, from equations (i) and equation (ii), we get,

2 ∠COA + 2 ∠COB = 180°

∠COA + ∠COB = 90°

Thus, ∠AOB = 90°

Hence proved.