Question

The common tangents AB and CD to two circles with centres O and O' intersect at E between their centres . Prove that the points O , E and O' are collinear .

Answer 1

Here Angle AEC and DEB are equal ( vertically opposite angles)

Join OA and OC,

So in triangle OAE and OCE, we have

OA = OC ( radii of same circle)

OE = OE (common)

$\angle$OAE = $\angle$OCE [90 each, as tangent is always perpendicular to its radius at point of contact]
$$\begin{gathered} △OEA\cong △OCE\left(\mathrm{RHS}\right) \\ So,\angle AEO=\angle CEO\left(\mathrm{CPCT}\right) \\ \mathrm{Similarly},\mathrm{for}\mathrm{the}\mathrm{other}\mathrm{circle}\mathrm{we}\mathrm{have} \\ \angle DEO\text{'}=\angle BEO\text{'} \\ \mathrm{Now}\angle AEC=\angle DEB \\ \Rightarrow \frac{1}{2}\angle AEC=\frac{1}{2}\angle DEB \\ \Rightarrow \angle AEO=\angle CEO=\angle DEO\text{'}=\angle BEO\text{'} \end{gathered}$$
So, all four angles are equal and bisected by OE and OE'.
Hence, O, E' and O' are collinear.