Question

In the below 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 angle AOB = 90°.​

Answer 1

Answer:

because anglesey was the hypotenuse triangle and every hypotenuse angles opposite side but that it was right angle

Answer 2

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

Similarly,

△OQB ≅ △OCB

So,

∠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°