Take a point $X$ on $CD$ such that $ XC = BC$. Now , $XB = CD - CX = CD - BC = AB = AD$ from the given conditions. Therefore ,$CBX$ , $ADX$ are both isosceles triangles with $CX = CB$ and $DX = DA$. Now , as $\triangle{ADB}$ is isosceles , $\angle{ABD} = \angle{ADB}$. From concyclicity , we have $\angle{BCA} = \angle{ADB} = \angle{ABD} = \angle{ACD}$. This implies that $CA$ is the angle bisector at $C$ of triangle $CBX$. Let $CA$ meet $BD$ at $Y$. Then , as $\triangle{CBX}$ is isosceles , $XY = YB$ and $\angle{XYC} = 90$. But , looking at $\triangle{ABX}$ , we can see that $\angle{AYX}= 90$ and $XY = YB$ which happens iff $\triangle{ABX} $ is isosceles with $AX = AB$. Therefore , $AX = AB$. But , $AX = AB = AD = DX$. Hence , $\triangle{ADX}$ is equilateral giving $\angle{CDA} = 60$.

Let $AB=a$ & $BC=b$. Extend $CD$ to $E$ such that $EC=BC$. Note that $\triangle EAC $ is A isosceles. So from Larry Noehn's Neglected formula , we get in $\triangle EAD$ , sides equal to $a,b,a^2+ab+b^2$.

Essentially the same as @2above .So to translate the sum condition , we mark a point $E$ on $BC$ such that $CE=CB$. Thus we get that $\triangle BCE$ and $\triangle AED$ are isosceles . Some angle chasing gives that $AC$ is the angle bisector and therefore also the perpendicular bisector in $\triangle BCE$ but in $\triangle ABE$ , the perpendicular bisector passes through the vertex $A$ and so $\triangle ABE$ is isosceles and so $\triangle AED $ is equilateral and so $\angle CDA=60$.