Question

Question 10
In a figure, the common tangents AB and CD to two circles with centres O and O’ intersect at E. Prove that the points O, E and O’ are collinear.

Answer 1

Joint AO, OC, O’D and O’B
In $\Delta$ EO’D and $\Delta$ EO’B
O’D = O’B [radius]
O’E = O’E [ common side]
ED = EB
[Since, tangents drawn from an external point to the circle are equal In length]

$\Delta E{O}^{\prime }D\cong \Delta E{O}^{\prime }B$ [by SSS congruence rule]
$\angle O^{\prime }ED=\angle O^{\prime }EB$
O'E is the angle bisector of $\angle$ DEB
Similarly, OE is the angle bisector of $\angle$ AEC.
Now, in quadrilateral DEBO',
$\angle O^{\prime }DE=\angle O^{\prime }BE={90}^{\circ }$ ...(i)
Since, AB is a straight line.
$\therefore \angle AED+\angle DEB={180}^{\circ }$ ...(ii)
$\Rightarrow \angle AED+{180}^{\circ }-\angle D{O}^{\prime }B={180}^{\circ }$
[From Eq. (ii)]
$\Rightarrow \angle AED=\angle D{O}^{\prime }B$ (iii)
Similarly $\angle AED=\angle AOC$ (iv)
Again from Eq. (ii) $\angle DEB={180}^{\circ }-\angle$ DO'B
Divided by 2 on both sides, we get
$\frac{1}{2}\angle DEB={90}^{\circ }-\frac{1}{2}\angle D{O}^{\prime }B$
$\Rightarrow \angle DE{O}^{\prime }={90}^{\circ }-\frac{1}{2}\angle D{O}^{\prime }B$ (v)
[Since, O'E is the angle bisector of $\angle DEBi.e.,\frac{1}{2}\angle DEB=\angle DE{O}^{\prime }$]
Similarly, $\angle AEC={180}^{\circ }-\angle AOC$
Divided by 2 on both sides, we get
$\frac{1}{2}\angle AEC={90}^{\circ }-\frac{1}{2}\angle AOC$
$\Rightarrow \angle AEO={90}^{\circ }-\frac{1}{2}\angle AOC$
[Since, OE is the angle bisector of $\angle$ AEC i.e., $\frac{1}{2}\angle AEC=\angle AEO$]
Now, $\angle AED+DE{O}^{\prime }+\angle AEO=\angle AED+({90}^{\circ }-\frac{1}{2}D{O}^{\prime }B)+({90}^{\circ }-\frac{1}{2}\angle AOC)$
$=\angle AED+{180}^{\circ }-\frac{1}{2}(\angle D{O}^{\prime }B+\angle AOC)$
$=\angle AED+{180}^{\circ }-\frac{1}{2}(\angle AED+\angle AED)$ [from eqs. (iii) and (iv)]
$=\angle AED+{180}^{\circ }-\frac{1}{2}(2\times \angle AED)$
$=\angle AED+{180}^{\circ }-\angle AED={180}^{\circ }$
$\therefore \angle AEO+\angle AED+\angle DE{O}^{\prime }={180}^{\circ }$
So. OEO; is straight line.
Dence O. E and O' are collinear
Hence proved