Firstly, it's clear that $ AC$ is the bisector of $ \angle BCD,$ because $ AB=AD.$ Let $ R$ be an internal point on $ CD$ such that $ BC=CR.$ Hence we have that $ \Delta ABC\cong \Delta ARC.$ From here $ AR=AB=AD$ and $ \angle ABC\cong \angle ARC.$ So we have to prove that $ \angle ARD<60^\circ.$ But from the condition $ CD>AB+BC$ we have that $ DR>AR=AD$ which means that $ \Delta ARD$ is isosceles with base bigger than his leg. Hence $ \angle ARD<60^\circ.$

Let $AB=AD=x, BC=y$ and $CD=x+z$ with $y < z$. Let $\theta = \angle ADC$ so $\angle ABC = 180^o-\theta$. It's easy to see that because $CD > BC$ then $\angle ABC > \angle ADC$ thus $0 < \theta < 90^o$ Now $AC^2 = x^2 + y^2 + 2xy \cos \theta = x^2 + (x+z)^2 - 2x(x+z)\cos \theta$, simplify it to get: $\cos \theta = \frac{(x+z)^2-y^2}{2x(x+y+z)} = \frac{x+z-y}{x}$ But since $\theta$ is an acute angle, $\theta < 60^o \iff \cos \theta > 1/2$ So we need to show $x+z-y > x$ which is true because $z > y$

CoBa_c_Kacka wrote: Firstly, it's clear that $ AC$ is the bisector of $ \angle BCD,$ because $ AB=AD.$ Let $ R$ be an internal point on $ CD$ such that $ BC=CR.$ Hence we have that $ \Delta ABC\cong \Delta ARC.$ From here $ AR=AB=AD$ and $ \angle ABC\cong \angle ARC.$ So we have to prove that $ \angle ARD<60^\circ.$ But from the condition $ CD>AB+BC$ we have that $ DR>AR=AD$ which means that $ \Delta ARD$ is isosceles with base bigger than his leg. Hence $ \angle ARD<60^\circ.$ Very nice solution. That was a brilliant introduction of R.