llanfairpg
It can be proved mathematically. I think. And I know I'm going to regret this.
Cd=Cd0+kCl^2
Divide through by Cl to give
(Cd/Cl) = Cd0/Cl + kCl
Assume, for illustrative purposes, that Cl = a . alpha (ie symmetric, away from stall - variable a is the slope of Cl vs alpha curve)
(Cd/Cl) = Cd0/(a.alpha) + k.a.alpha
(Cd/Cl) is a minimum with respect to alpha (ie Cl/Cd is maximum) when you differentiate (Cd/Cl) w.r.t alpha and set the result to zero.
Do this and you get k(a.alpha)^2 = Cd0 or kCl^2 = Cd0 at best Cl/Cd.
But lo and behold....kCl^2 = Cd0 is exactly the condition at minimum drag. (This is intuitive but can also be proved mathematically). QED
Where's Genghis when you need him....