In figure, the common tangent, AB and CD to two circle with centers O and O' intersects at E.?

You basically have to prove that the angle OEO' is 180 degree:
First join O to A and C and O' to D and B.
And for convinience, let angle AOE be x.
Now since angle OAE is 90, AEO is (90-x)
From figure, DEO' is also (90-x)
Now consider triangle OCE. It's exterior angle OED will be equal to (90+x)
If you subtract the angle AEO which is (90-x) from OED which is (90+x), angle AED will be 2x.
Finally, angle OEO' is the sum of AEO, OED and DEO' which will come out to be 180.
So te points are colinear.

I like hints, so here you go:
AE = CE and DE = EB