the common tangent ab and cd to two circles with centres o and o' intersect at E.Prove that points o,e,o' are collinear.

the tangents are between the circle and not like other figures explained by meritnation experts.the fig. is diff

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]

$∆\mathrm{OAE}\cong ∆\mathrm{OCE} \left(\mathrm{RHS}\right)$

So Angle AEO = angle CEO  (CPCT)

Similarly for other circle, we get

Angle DEO' = Angle BEO'

$$\begin{gathered} \mathrm{Now}, \angle \mathrm{AEC} = \angle \mathrm{DEB} \\ \Rightarrow \frac{1}{2}\angle \mathrm{AEC} = \frac{1}{2}\angle \mathrm{DEB} \\ \Rightarrow \angle \mathrm{AEO} = \angle \mathrm{CEO} = \angle \mathrm{DEO}\text{'} = \angle \mathrm{BEO}\text{'} \end{gathered}$$

So all four angle are equal and bisected by OE and O'E.

Hence O ,E' and O are collinear.