In Figure 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°.
From the figure given in the textbook, join OC. Now, the diagram will be as
Now the triangles △OPA and △OCA are similar using SSS congruency as:
(i) OP = OC They are the radii of the same circle
(ii) AO = AO It is the common side
(iii) AP = AC These are the tangents from point A
So, △OPA ≅ △OCA
△OQB ≅ △OCB
∠POA = ∠COA … (Equation i)
And, ∠QOB = ∠COB … (Equation ii)
Since the line POQ is a straight line, it can be considered as a 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°
∴∠AOB = 90°