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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, is intersecting XY at A and X'Y' at B. Prove that $\angle AO\!B=90^{\circ}$ .

Answers (1)

Given :

XY is tangent at P.

X'Y' is tangent at Q.

And $XY\parallel X'Y'$

AB is tangent at C.

To prove : $\angle AOB=90^{\circ}$

Proof :

Theorem 1 $\rightarrow$ Tangent at any point of circle is perpendicular to the radius through point of contact.

For tangent AB and radius OC

$OC\perp AB$ (from theorem 1)

So, $\angle ACO =\angle BCO =90^{\circ}$

In $\bigtriangleup AOP$ and $\bigtriangleup AOC$

$OP=OC$ (both are radius)

$AP=AC$ ( by theorem 1)

$OA=OA$ (common)

$\therefore \bigtriangleup AOC \cong AOP$ (SSS congruency rule)

Hence by CPCT, $\angle AOC = \angle AOP\; \; \; \; -(i)$

Now in $\bigtriangleup BOC \; \; \&\; \; \bigtriangleup BOQ$

$OC=OQ$ ( both are radius)

$BC=BQ$ ( by theorem 1)

$OB=OB$ ( common )

$\therefore \bigtriangleup BOC \cong \bigtriangleup BOQ$ (by SSS congruency rule)

Hence by CPCT, $\angle BOC=\angle BOQ\; \; \; -(ii)$

$\rightarrow$ For line PQ,

$\angle AOP +\angle AOC +\angle BOC +\angle BOQ=180^{\circ}$

$\Rightarrow \angle AOC +\angle AOC +\angle BOC +\angle BOC=180^{\circ}$ (from (i) and (ii))

$\Rightarrow 2\angle AOC +2\angle BOC =180^{\circ}$

$\Rightarrow \angle AOC +\angle BOC =90^{\circ}$

$\Rightarrow \angle AOB =90^{\circ}$

Hence proved.

Posted by

Ravindra Pindel

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