In a figure the common tangents, AB and CD to two circles with centres 0 and O’ intersect at E. Prove that the points 0, E and O’ are collinear.
Joint $A O, O C$ and $O^{\prime} D, O^{\prime} B$.
Now, in $\triangle E O^{\prime} D$ and $\triangle E O^{\prime} B$,
$O^{\prime} D=O^{\prime} B$ [radius]
$O^{\prime} E=O^{\prime} E$ [common side]
$E D=E B$
[since, tangents drawn from an external point to the circle are equal in length]
$\therefore$ $\triangle E O^{\prime} D \cong \triangle E O^{\prime} B$ [by SSS congruence ruie]
$\Rightarrow$ $\angle O^{\prime} E D=\angle O^{\prime} E B$ $\ldots$ (i)
$O^{\prime} E$ is the angle bisector of $\angle D E B$.
Similarly, $O E$ is the angle bisector of $\angle A E C$.
Now, in quadrilateral $D E B O^{\prime}$,
$\angle O^{\prime} D E=\angle O^{\prime} B E=90^{\circ}$
[since, $C E D$ is a tangent to the circle and $O^{\prime} D$ is the radius, $i, \theta, O^{\prime} D \perp C E D$ ]
$\Rightarrow \quad \angle O^{\prime} D E+\angle O^{\prime} B E=180^{\circ}$
$\therefore \quad \angle D E B+\angle D O^{\prime} B=180^{\circ} \quad$ [since, $D E B O^{\prime}$ is cyclic quadrilateral] ...(ii)
Since, $A B$ is a straight line.
$\therefore \quad \angle A E D+\angle D E B=180^{\circ}$
$\Rightarrow \quad \angle A E D+180^{\circ}-\angle D O^{\prime} B=180^{\circ} \quad$ [from Eq. (ii)]
$\Rightarrow \quad \angle A E D=\angle D O^{\prime} B$ .....(iii)
Similarly, $\angle A E D=\angle A O C$ ...(iv)
Again from Eq. (ii), $\quad \angle D E B=180^{\circ}-\angle D O^{\prime} B$
Divided by 2 on both sides, we get
$\frac{1}{2} \angle D E B=90^{\circ}-\frac{1}{2} \angle D O^{\prime} B$
Divided by 2 on both sides, we get
$\frac{1}{2} \angle D E B=90^{\circ}-\frac{1}{2} \angle D O^{\prime} B$
$\Rightarrow$ $\angle D E O^{\prime}=90^{\circ}-\frac{1}{2} \angle D O^{\prime} B$ ....$(\mathrm{v})$
[since, $O^{\prime} E$ is the angle bisector of $\angle D E B i . e, \frac{1}{2} \angle D E B=\angle D E O$ ]
Similarly, $\angle A E C=180^{\circ}-\angle A O C$
Divided bv 2 on both sides, we get
$\frac{1}{2} \angle A E C=90^{\circ}-\frac{1}{2} \angle A O C$
$\Rightarrow$ $\angle A E O=90^{\circ}-\frac{1}{2} \angle A O C$ ....(vi)
[since, $O E$ is the angle bisector of $\angle A E C$ i.e., $\frac{1}{2} \angle A E C=\angle A E O$ ]
Now, $\angle A E D+\angle D E O^{\prime}+\angle A E O=\angle A E D+\left(90^{\circ}-\frac{1}{2} \angle D O^{\prime} B\right)+\left(90^{\circ}-\frac{1}{2} \angle A O C\right)$
$=\angle A E D+180^{\circ}-\frac{1}{2}\left(\angle D O^{\prime} B+\angle A O C\right)$
$=\angle A E D+180^{\circ}-\frac{1}{2}(\angle A E D+\angle A E D)$ [from Eqs. (iii) and (iv)]
$=\angle A E D+180^{\circ}-\frac{1}{2}(2 \times \angle A E D)$
$=\angle A E D+180^{\circ}-\angle A E D=180^{\circ}$
$\therefore \quad \angle A E O+\angle A E D+\angle D E O^{\prime}=180^{\circ}$
So, OEO' is straight line.
Hence, $O, E$ and $O^{\prime}$ are collinear.