In figure, the common tangent, \(AB\) and \(CD\) to two circles with centre \(O\) and \(O'\) intersect at \(E\). Prove that the points \(O\), \(E\), \(O'\) are collinear.
Proof:
\(AB\) and \(CD\) are two tangents with centres \(O\) and \(o'\) intersect at \(E\).
Join \(AO\), \(OC\), \(O'D\) and \(O'B\).
In \(\Delta OAE\) and \(\Delta OCE\),
\(OA = \) (radii of the same circle)
\(OE = OE\) (common side)
\(\angle OAE = \angle\)
Thus, by congruence criterion, \(\Delta OAE \cong \Delta OCE\)
We know that, orresponding parts of congruent triangles are congruent.
\(\angle AEO = \angle CEO\)
Similarly, \(\angle BEO' = \angle DEO'\)
\(\angle AEC = \angle \)
\(\frac{1}{2} \angle AEC = \frac{1}{2} \angle DEB\)
\(\angle AEO = \angle CEO = \angle BEO' = \angle DEO'\)
Since these angles are equal and are bisected by and \(O'E\),
\(O\), \(E\), \(O'\) are collinear.