Name $\angle BAD = \alpha$, $\angle BCD = 2\alpha$, $\angle ADC = \beta$, $\angle ABC = 2\beta$. The interior angles of $ABCD$ add up to $3\alpha + 3\beta = 360^\circ$. Hence, $\alpha + \beta = 120^\circ$. As $\alpha \neq \beta$, we may assume without loss of generality that $\beta < 60^\circ < \alpha$.

Let $X = AB \cap CD$. Notice that $\angle AXD = 60^\circ$. Let $E$ be the point on the same side as $A$ with respect to line $CD$ such that $CDE$ is equilateral. We have $\angle ECD = 60^\circ = \angle AXD$ and $CE = CD = AB$. Hence, $EC \parallel AB$ and $ABCE$ is a parallelogram. This implies $\angle BAE = \angle BCE = 2\alpha - 60^\circ$ and $\angle EAD = (2\alpha - 60^\circ) - \alpha = \alpha - 60^\circ$. On the other hand, $\angle EDA = 60^\circ - \beta$. Since $60^\circ - \beta = \alpha - 60^\circ > 0$, triangle $AED$ is isosceles with $AE = ED$. Hence, $ABC$ and $BCD$ are also isosceles. In particular $\angle BAC = \frac{180^\circ - \angle ABC}{2} = 90^\circ - \beta$, which implies $\angle CAD = \alpha - (90^\circ - \beta) = 30^\circ$; and $\angle BDC = \frac{180^\circ - \angle BCD}{2} = 90^\circ - \alpha$, which implies $\angle BDA = \beta - (90^\circ - \alpha) = 30^\circ$. Therefore, the acute angle formed by $AC$ and $BD$ is $30^\circ + 30^\circ = 60^\circ$.