PQ and RS are two parallel tangents to a circle with centre'O',another tangent AB with the point of contact 'C' intersecting PQ and RS at A and B respectively.Prove that angle AOB=90 degree.

If your question is -

PQ and RS are two parallel tangents to a circle with centre 'O ',another tangent** XY **with the point of contact 'C ' intersecting PQ and RS at A and B respectively. Prove that angle AOB=90 degree.

Then the solution is as follows:

Construction: Join O to A, O to B and O to C.

Since tangent is perpendicular to the radius through the point of contact.

∴∠x=∠y=90°

Now,

In right ΔOMA and ΔOCA.

Hyp. OA = Hyp. OA [common]

OM= OC [radii of the same circle]

∠OMA=∠OCA [each equal to 90°]

⇒ΔOMA ΔOCA[By RHS congruence criterion]

⇒∠1=∠2 [By Cpct]

Similarly, ∠3=∠4

Now,

∠1+∠2+∠3+∠4=180° [Linear pair]

⇒2(∠2+∠3)=180°

⇒∠2+∠3=90°

∴∠AOB=90°

Hence Proved.

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