I3.3 In the figure, triangle $A B C$ has $\angle A B C=2 \beta^{\circ}, A B=A D$ and $C B=C E$. If $\gamma^{\circ}=\angle D B E$, determine the value of $\gamma$.
Let $\angle A B E=x$
$$
\begin{array}{l}
\angle A B C=90^{\circ} \\
\angle C B E=90^{\circ}-x \\
\angle A D B=x+\gamma^{\circ} \text { (base } \angle \mathrm{s} \text { isos. } \triangle \text { ) } \\
\angle C E B=\angle C B E=90^{\circ}-x \text { (base } \angle \mathrm{s} \text { isos. } \Delta \text { ) } \\
\text { In } \triangle B D E, \gamma^{\circ}+x+\gamma^{\circ}+90^{\circ}-x=180^{\circ}(\angle \mathrm{s} \text { sum of } \Delta) \\
\gamma=45
\end{array}
$$