Question

Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

**OR**

A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD = AD + BC.

Solution

Given: AB is tangent to a circle with centre O.

To prove: OP is perpendicular to AB.

Constructions: Take a point Q on AB and join OQ.

Proof: Since Q is a point on the tangent AB, other than the point of contact P, so Q will be outside the circle.

Let OQ intersect the circle at R.

Now OQ = OR + RQ

$\Rightarrow \mathrm{OQ}\mathrm{OR}\Rightarrow \mathrm{OQ}\mathrm{OP}.......[\mathrm{as}\mathrm{OR}=\mathrm{OP}]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}\mathrm{OQ}$

Thus OP is shorter than any other segment among all and the shortest length is the perpendicular from O on AB.

$\therefore \mathrm{OP}\perp \mathrm{AB}.$ Hence proved.

**OR**

Let ABCD be a quadrilateral, circumscribing a circle.

Since the tangents drawn to the circle from an external point are equal,

we have,

AP = AS ..........(1)

BP = BQ ..........(2)

RC = QC ..........(3)

DR = DS ..........(4)

Adding, (1), (2), (3) and (4), we get

AP + PB + RC + DR = AS + BQ + QC + DS

(AP + PB ) + (RC + DR ) = (AS + DS ) + (BQ + QC)

AB + CD = AD + BC.

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Fig. 9.12.

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Fig. 9.13.

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