$\underline{\textbf{Solution}}$ Because $\angle{A}=45^\circ$, it is possible to flip both $\triangle{DAC}$ and $\triangle{DAB}$ outwards to create a right angle from which we can construct a square as shown.

Let $AD=x$, then applying the Pythagorean theorem on the bottom right triangle leads $$(x-3)^2 + (x-2)^2 = 5^2\implies x=6\implies S_\triangle{ABC}=\frac{1}{2}\times 5\times 6=\boxed{15}$$ $\underline{\textbf{Alternative Solution}}$ Construct the other two altitudes $BE$ and $CF$, let the othocenter be $H$. It follows that $(B, D, H, F)$, $(C, D, H, E)$ and $(B, C, E, F)$ are all concyclic. $\angle{BAC} = 45^\circ \implies \angle{ABH}=\angle{ACH}=\angle{FDH}= \angle{ADF} = \angle{ADE} = 45^\circ$. If follows that $DF$ and $DE$ bisects $\angle{ADB}$ and $\angle{ADC}$, respectively. Let $AD=h$, $AB=c$, and $AC=b$. Hence, we have $\frac{AD}{DC}=\frac{h}{2}=\frac{AE}{EC}=\frac{\frac{\sqrt{2}}{2}c}{b-\frac{\sqrt{2}}{2}c} \implies \frac{h}{2+h}=\frac{\frac{\sqrt{2}}{2}c}{b}$. Similiarly, we have $\frac{h}{3+h}=\frac{\frac{\sqrt{2}}{2}b}{c}$. Hence $\frac{h}{2+h}\cdot\frac{h}{3+h}=\frac{1}{2} \implies h = 6 \implies S_{\triangle{ABC}}=\boxed{15}$.