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

#### Solution

Join AO, OC and O’D, O’B.

Now, in ∆EO’D and ∆EO’B,

O’D = O’B

O’E = O’E

ED = EB .......[Tangents drawn from an external point to the circle are equal in length

∴ EO’D ≅ ∆ EO’B .......[By SSS congruence criterion]

⇒ ∠O’ED = ∠O’EB ......(i)

i.e., O’E is the angle bisector of ∠DEB.

Similarly, OE is the angle bisector of ∠AEC.

Now, in quadrilateral DEBO’.

∠O’DE = ∠O’BE = 90° .......[CED is a tangent to the circle and O’D is the radius, i.e., O’D ⊥ CED]

⇒ ∠O’DE + ∠O’BE = 180°

∴ ∠DEB + ∠DO’B = 180° .......[∵ DEBO’ is cyclic quadrilateral] .......(ii)

Since, AB is a straight line.

∴ ∠AED + ∠DEB = 180°

⇒ ∠AED + 180° – ∠DO’B = 180° .....[From (ii)]

⇒ ∠AED = ∠DO’B ......(iii)

Similarly, ∠AED = ∠AOC .......(iv)

Again from eq. (ii), ∠DEB = 180° – ∠DO’B

Dividing by 2 on both sides, we get

`1/2 ∠DEB = 90^circ - 1/2` ∠DO'B

⇒ ∠DE'O = 90° `- 1/2` ∠DO'B ......(v) [∵ O'E is the angle bisector of ∠DEB i.e. `1/2` ∠DEB = ∠DEO']

Similarly, ∠AEC = 180° – ∠AOC

Dividing vy 2 on both sides, we get

`1/2 ∠AEC = 90^circ - 1/2 ∠AOC` .....(vi) [∵ OE is the angle bisector of ∠AEC i.e., `1/2 ∠AEC = ∠AEO`]

Now, ∠AED + ∠DEO' + ∠AEO = ∠AED + `(90^circ - 1/2 ∠DO"'"B) + (90^circ - 1/2 ∠AOC)`

= `∠AED + 180^circ - 1/2 (∠DO"'"B + ∠AOC)`

= `∠AED + 180^circ - 1/2 (∠AED + ∠AED)` .....[From equation (iii) and (iv)]

= `∠AED + 180^circ - 1/2 (2 xx ∠AED)`

= ∠AED + 180° – ∠AED = 180°

∴ ∠AED + ∠DEO' + ∠AED = 180°

So, OEO’ is straight line.

Hence, O, E and O’ are collinear.