Question

In the given 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°.​

Answer 1

Step-by-step explanation:

Given, XY and MN are two parallel tangents to a circle with centre O

and another tangent AB with the point of contact C intersecting XY at A and MN at B.

Now, join O to A, O to B and O to C.

Since the tangents are perpendicular to the radius through the point of contact.

So, ∠x = ∠y = 90

Now, in the triangle OMA and OCA,

OA = OA {common}

OM = OC {radii of the circle}

∠OMA = ∠OCA {each 90}

By SSA congrunec criterian,

ΔOMA ≅ ΔOCA

So, ∠1 = ∠2 {by CPCT}

Similarly, ∠3 = ∠4

Now, ∠1 + ∠2 + ∠3 + ∠4 = 180

=> 2(∠2 + ∠3) = 180

=> ∠2 + ∠3 = 180/2

=> ∠2 + ∠3 = 90

=> ∠AOB = 90

Hence proved.