Solution:

Let $P$ be the reflection of $C$ with respect to the line $AB$. From the congruence of triangles $ABC$ and $ABP$ we obtain that
1. $A\hat{B}P=A\hat{B}C=\delta+\epsilon$;
2. $A\hat{P}B=A\hat{C}B=\gamma$;
3. $AP=AC$ and $BP=BC$.

Using the first piece of information we have that
$$D\hat{B}P=D\hat{B}A+A\hat{B}P=\epsilon+(\delta+\epsilon)=\delta+2\epsilon=180^\circ$$
and therefore the points $P$, $B$, $D$ are collinear.

Now consider the triangle $APD$. Using the second piece of information we have that
$$A\hat{P}D+A\hat{D}P=A\hat{P}B+A\hat{D}B=\gamma+\beta=90^\circ$$
hence the triangle $APD$ is right-angled at $A$.
$$(DB+BC)^2=(DB+BP)^2=DP^2=AP^2+AD^2=AC^2+AD^2,$$
which is what we wanted to prove.