Question

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.

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

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°  ...(ii) [∵ DEBO’ is cyclic quadrilateral]

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 equation (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 by 2 on both sides, we get

`1/2 ∠"AEC" = 90^circ - 1/2 ∠"AOC"`

⇒ `∠"AEO" = 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' + ∠AEO = 180°

So, OEO’ is straight line.

Hence, O, E and O’ are collinear.