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Home » 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°.
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°.
CBSE | Circles | Class 10 | Maths | NCERT
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°.

Answer:

From the given figure in the textbook, join OC. The diagram will now be as-

Now using the SSS congruency the triangles △OPA and △OCA are similar:

(i) OP = OC since they are the same circle’s radii

(ii) AO = AO as it is the common side

(iii) AP = AC as these are the tangents from point A

As a result, △OPA ≅ △OCA

In the same way,

△OQB ≅ △OCB

As a result,

∠POA = ∠COA … (Eq. (i))

And, ∠QOB = ∠COB … (Eq. (ii))

Because the line POQ is a straight line, it can be regarded as a circle’s diameter.

As a result, ∠POA +∠COA +∠COB +∠QOB = 180°

Now, from eq. (i) and eq. (ii) we get,

2∠COA+2∠COB = 180°

∠COA+∠COB = 90°

Therefore, ∠AOB = 90°

i

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