In the figure,the common tangents $AB$ and $CD$ to two circles with centers $O$ and $O^{\prime}$ intersect at $E$. Prove that the points $O, E, O^{\prime}$ are collinear.

(N/A) Join $AO, OC$ and $O^{\prime}D, O^{\prime}B$.
In $\triangle E O^{\prime} D$ and $\triangle E O^{\prime} B$:
$O^{\prime} D = O^{\prime} B$ (radii of the same circle)
$O^{\prime} E = O^{\prime} E$ (common side)
$ED = EB$ (tangents drawn from an external point $E$ to the circle are equal in length)
Therefore,$\triangle E O^{\prime} D \cong \triangle E O^{\prime} B$ (by $SSS$ congruence rule).
This implies $\angle O^{\prime} E D = \angle O^{\prime} E B$,so $O^{\prime} E$ is the angle bisector of $\angle DEB$.
Similarly,by considering $\triangle EOA$ and $\triangle EOC$,we can prove $\triangle EOA \cong \triangle EOC$,which implies $\angle OEA = \angle OEC$,so $OE$ is the angle bisector of $\angle AEC$.
Since $AB$ and $CD$ are straight lines intersecting at $E$,$\angle AEC$ and $\angle DEB$ are vertically opposite angles,so $\angle AEC = \angle DEB$.
Since $OE$ bisects $\angle AEC$ and $O^{\prime}E$ bisects $\angle DEB$,and $\angle AEC = \angle DEB$,it follows that $\angle OEC = \angle O^{\prime}EB$.
Since $C, E, B$ lie on a straight line,$\angle OEC + \angle OEO^{\prime} + \angle O^{\prime}EB = 180^{\circ}$.
Substituting $\angle OEC = \angle O^{\prime}EB$,we get $2\angle O^{\prime}EB + \angle OEO^{\prime} = 180^{\circ}$.
However,since $O, E, O^{\prime}$ form a straight line,the angle $\angle OEO^{\prime}$ must be $180^{\circ}$.
Thus,$O, E, O^{\prime}$ are collinear.