Question

The common tangents AB and CD to two circles with centers O and O' intersect at E between their centers. Prove that the points O, E and O' are collinear,

Answer 1

∠AEC = ∠DEB (Vertically Opposite Angle)

Join OA and OC

so, in triangle OAE and triangle OCE we have,

OA=OC (radii of same circle)

OE=OE (common)

∠OAE=∠OCE (90° as the tangent is always perpendicular to the radius at the point of contact)

∴ ΔOAE ≡ ΔOCE

so ∠AEO = ∠CEO (CPCT)

similarly for other circle we have,

∠DEO' = ∠BEO' (CPCT)

Now ∠AEC=∠DEB

⇒1/2(∠AEC) = 1/2(∠DEB)

⇒∠AEO =∠CEO = ∠DEO'= ∠BEO'

so all 4 ∠'s are equal and bisected by OE and OE'

∴ O,E,O' are collinear