In the figure, two equal circles, with centres O and O', touch each other at X. O'X produced meets the circle with centre O at A. AC is tangent to the circle with centre O, at the point C. O'D is perpendicular to AC. Find the value of DO'CO.

O'D is perpendicular to AC. We know that ∠ADO' = 90°
Radius OC is perpendicular to tangent AC.
In △ADO' and △ACO,
∠ADO' = ∠ACO (each 90 degree)
∠DAO = ∠CAO (common)
By AA property, triangles ADO' and ACO are similar to each other.
AO'/AO = DO'/CO (corresponding sides of similar triangles)
AO = AO' + O'X + OX = 3AO' (Since AO' = O'X = OX because radii of the two circles are equal)
AO'/AO = AO'/3AO' = 1/3
DO'/CO = AO'/AO = 1/3
DO'CO = 1/3.