Question

In figure, XY and X’Y’ are two parallel tangents to a circle , x 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°.

Answer 1

Answer:

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

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°

Answer 2

Answer:

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

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°

Step-by-step explanation:

POA +∠COA +∠COB +∠QOB = 180°