Problem: Let $ABC$ be a triangle with $\angle A = 90\dg$.
Points $D$ and $E$ lie on sides $AC$ and $AB$, respectively,
such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$.
Segments $BD$ and $CE$ meet at $I$.
Determine whether or not it is possible for segments $AB$, $AC$, $BI$, $ID$, $CI$, $IE$
to all have integer lengths.

Solution: The answer is no.
We prove that it is not even possible that $AB$, $AC$, $CI$, $IB$ are all integers.

\begin{center}
  \begin{asy}
    size(5cm);
    pair B = dir(140);
    pair A = conj(B);
    pair C = -B;
    filldraw(A--B--C--cycle, opacity(0.1)+lightblue, blue);
    pair I = incenter(A, B, C);
    pair D = extension(B, I, A, C);
    pair E = extension(C, I, A, B);
    draw(B--D);
    draw(C--E);
    filldraw(incircle(A,B,C), opacity(0.1)+lightred, red+dotted);

    dot("$B$", B, dir(B));
    dot("$A$", A, dir(A));
    dot("$C$", C, dir(C));
    dot("$I$", I, dir(45));
    dot("$D$", D, dir(D));
    dot("$E$", E, dir(E));

    /* Source generated by TSQ */
  \end{asy}
\end{center}

First, we claim that $\angle BIC = 135\dg$.
To see why, note that
\[ \angle IBC + \angle ICB = \frac{\angle B}{2} + \frac{\angle C}{2}
  = \frac{90\dg}{2} = 45\dg. \]
So, $\angle BIC = 180\dg - (\angle IBC + \angle ICB) = 135\dg$, as desired.

We now proceed by contradiction.
The Pythagorean theorem implies
\[ BC^2 = AB^2 + AC^2 \]
and so $BC^2$ is an integer.
However, the law of cosines gives
\begin{align*}
  BC^2 &= BI^2 + CI^2 - 2 BI \cdot CI \cos \angle BIC \\
  &= BI^2 + CI^2 + BI \cdot CI \cdot \sqrt 2.
\end{align*}
which is irrational, and this produces the desired contradiction.