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In Figure, the common tangent, AB and CD to two circles with centres O and O' intersect at E. Prove that the points O, E, O' are collinear.

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Solution
Construction : Join AO and OS
${O}'D$ and ${O}'B$

In $\bigtriangleup E{O}'B$ and $\bigtriangleup E{O}'D$
${O}'D= {O}'B$    [Radius are equal]
${O}'E= {O}'E$ [Common side]
ED = EB           [Tangent drawn from an external point to the line circle are equal to length]
$AE{O}'B\cong E{O}'D$   [By SSS congruence criterion]
$\Rightarrow \; \angle {O}'ED= \angle {O}'EB\, \cdots \left ( i \right )$
i.e.,      ${O}'E$ is bisector of $\angle BED$
Similarly OE is bisector of $\angle AEC$
In quadrilateral $DEB{O}'$
$\angle {O}'DE= \angle {O}'BE= 90^{\circ}$
$\Rightarrow \, \angle {O}'DE= \angle {O}'BE= 180^{\circ}$
$\therefore \angle DEB+\angle D {O}'B= 180^{\circ}\, \cdots \left ( ii \right )$ [ $\mathbb{Q}DEB{O}'$ is cyclic quadrilateral]
$\angle AED+\angle DEB= 180^{\circ}$ [ $\mathbb{Q}$ AB is a straight line]
$\angle AED+180^{\circ}-\angle D{O}'B= 180^{\circ}$ [From equation (ii)]
$\angle AED-\angle D{O}'B= 0$
$\angle AED-\angle D{O}'B\: \: \cdots \left ( iii \right )$
Similarly $\angle AED= \angle ADC\: \:\cdots \left ( iv \right )$
From equation (ii) $\angle DEB= 180^{\circ}-\angle D{O}'B$
Dividing both side by 2
$\frac{\angle DEB}{2}= \frac{180^{\circ}-\angle D{O}'B}{2}$
$\angle DE{O}'= 90^{\circ}-\frac{1}{2}\angle D{O}'B\: \: \cdots \left ( v \right )$
$\left [ \because \, \frac{\angle DEB}{2}= \angle DE{O}' \right ]$
Similarly $\angle AEC= 180^{\circ}-\angle AOC$
Dividing both side by 2
$\frac{1}{2}\angle AEC= 90^{\circ}-\frac{\angle AOC}{2}$
$\angle AEO= 90^{\circ}-\frac{1}{2}\angle AOC\; \cdots \left ( vi \right )$
$\left [ \because \frac{1}{2}\angle AEC= \angle AEO \right ]$
Now $\angle AEO+\angle AED+\angle DE{O}'$
$= 90-\frac{1}{2}\angle AOC+\angle AED+90^{\circ}-\frac{1}{2}\angle D{O}'B$
$= \angle AED+180^{\circ}-\frac{1}{2}\left ( \angle D{O}'B+\angle AOC \right )$
$= \angle AED+180^{\circ}-\frac{1}{2} \left [ \angle AED+\angle AED \right ]$ [from equation (iii) and (iv)]
$= \angle AED+180^{\circ}-\frac{1}{2} \left [ 2\angle AED \right ]$
$\angle AED+180^{\circ}-\angle AED$
$= 180^{\circ}$
$\therefore \: \angle AED+\angle DE{O}'+\angle AEO= 180^{\circ}$
So,OEO’ is straight line
$\therefore$ O, E and ${O}'$ are collinear.
Hence Proved