In acute triangle $ABC,$ $\angle A = 45^\circ.$  Let $D$ be the foot of the altitude from $A$ to $\overline{BC}.$  if $BD = 2$ and $CD = 3,$ then find the area of triangle $ABC.$
Let $x = AD.$

[asy]
unitsize(0.5 cm);

pair A, B, C, D;

A = (2,6);
B = (0,0);
C = (5,0);
D = (2,0);

draw(A--B--C--cycle);
draw(A--D);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$2$", (B + D)/2, S, red);
label("$3$", (C + D)/2, S, red);
label("$x$", (A + D)/2, E, red);
[/asy]

Then $\tan \angle BAD = \frac{2}{x}$ and $\tan \angle CAD = \frac{3}{x}.$

We know that $\angle BAC = 45^\circ.$  By the tangent addition formula,
\begin{align*}
\tan \angle BAC &= \tan (\angle BAD + \angle CAD) \\
&= \frac{\tan \angle BAD + \tan \angle CAD}{1 - \tan \angle BAD \tan \angle CAD} \\
&= \frac{2/x + 3/x}{1 - 2/x \cdot 3/x} \\
&= \frac{5x}{x^2 - 6}.
\end{align*}Then $5x = x^2 - 6,$ or $x^2 - 5x - 6 = 0.$  This factors as $(x - 6)(x + 1) = 0,$  so $x = 6.$  The area of triangle $ABC$ is then $\frac{1}{2} \cdot 6 \cdot 5 = \boxed{15}.$