Angle in a triangle with equal segments

Let $\angle A = \alpha$ and $\angle B = \beta$. In $\triangle ABC$, $\alpha + \beta + 50^\circ = 180^\circ \implies \alpha + \beta = 130^\circ$.

In $\triangle ADE$, since $AD = DE$, it is an isosceles triangle with $\angle AED = \angle DAE = \alpha$. Thus, $\angle ADE = 180^\circ - 2\alpha$.

In $\triangle BDF$, since $BD = DF$, it is an isosceles triangle with $\angle DFB = \angle DBF = \beta$. Thus, $\angle BDF = 180^\circ - 2\beta$.

Since points A, D, B lie on a straight line: $\angle BDF + \angle FDE + \angle ADE = 180^\circ$ $(180^\circ - 2\beta) + \angle FDE + (180^\circ - 2\alpha) = 180^\circ$ $\angle FDE = 2(\alpha + \beta) - 180^\circ = 2(130^\circ) - 180^\circ = 260^\circ - 180^\circ = 80^\circ$.