In a figure the common tangents, AB and CD to two circles with centers O and O' intersect at E. Prove that the points O, E and O' are collinear.

Explanation

Joint AO, OC and O'D,O'B.
Now, in`DeltaEO'Dand DeltaEO'B,`
O'D=O'B [radius] O'E=O'E [common side]
ED=EB
[since, tangents drawn from an external point to the circle are equal in length]

`:.DeltaEO'D~=DeltaEO'B` [by SSS congruence rule]
`impliesangleO'ED=angleO'EB`
O'E is the angle bisector of `angleDEB.....(i)`
Similarly, OE is the angle bisector of `angleAEC.`
Now, in quadrilateral DEBO'.
`angleO'DE=angleO'BE=90^(@)`
[since, CED is a tangent to the circle and O'D is the radius, i.e., `O'DbotCED`]
`rArrangleO'DE+angleO'BE=180^(@)`
`:.angleDEB+angleDO'B=180^(@)` [since, DEBO' is cyclic quadrilateral]......(ii)
Since, AB is a straight line.
`:.angleAED+angleDEB=180^(@)`
`impliesangleAED+180^(@)=angleDO'B=180^(@) [from Eq. (ii)]`
`impliesangleAED=angleDO'B.....(iii)`
Similarly, `angleAED=angleAOC.....(iv)`
Again from Eq. (ii), `angleDEB=180^(@)-angleDO'B`
Divided by 2 on both sides, we get
`(1)/(2)angleDEB=90^(@)-(1)/(2)angleDO'B`
`impliesangleDEO'=90^(@)-(1)/(2)angleDO'B.....(v)`
[since, O'E is the angle bisector of `angleDEB" "i.e.,(1)/(2)angleDEB=angleDEO`]
Similary, `angleAEC=180^(@)-angleAOC`
Divided by 2 on both sides, we get
`(1)/(2)angleAEC=90^(@)-(1)/(2)angleAOC`
`impliesangleAEO=90^(@)-(1)/(2)angleAOC......(vi)`
[since, OE is the angle bisector of `angleAEC" "i.e.,(1)/(2)angleAEC=angleAEO`]
Now, `angleAED+angleDEO+angleAEO+angleAED+(90^(@)-(1)/(2)angleDO'B)+(90^(@)-(1)/(2)angleAOC)`
`impliesangleAED+180^(@)-(1)/(2)(angleDO'B+angleAOC)`
`impliesangleAED+180^(@)-(1)/(2)(angleAED+angleAED)` [from Eqs. (iii) and (iv)]
`impliesangleAED+180^(@)-(1)/(2)(2xxangleAED)`
`impliesangleAED+180^(@)-angleAED=180^(@)`
`:.angleAEO+angleAED+angleDEO'+180^(@)`
So, OEO' is straight line.
Hence, O, E and O' are collinear. Hence proved.