Problem: Let $ABCD$ be a rhombus, and let $K$ and $L$ be points
such that $K$ lies inside the rhombus, $L$ lies outside the rhombus,
and $KA = KB = LC = LD$.
Prove that there exist points $X$ and $Y$ on lines $AC$ and $BD$
such that $KXLY$ is also a rhombus.

Solution: To start, notice that $\triangle AKB \cong \triangle DLC$ by SSS.
Then by the condition $K$ lies inside the rhombus while $L$ lies outside it,
we find that the two congruent triangles are just translations of each other
(i.e.\ they have the same orientation).

\paragraph{First solution.}
Let $M$ be the midpoint of $\ol{KL}$ and is $O$ the center of the rhombus.
\begin{claim*}
  $\ol{MO} \perp \ol{AB}$.
\end{claim*}
\begin{proof}
  Let $U$ and $V$ denote the midpoint of $\ol{AB}$ and $\ol{CD}$ respectively.
  Then $\ol{KU}$ and $\ol{LV}$ are obviously translates,
  and perpendicular to $\ol{AB} \parallel \ol{CD}$.
  Since $M$ is the midpoint of $\ol{KL}$ and $O$ is the midpoint of $\ol{UV}$,
  the result follows.
\end{proof}

We choose $X$ and $Y$ to be the intersections of
the perpendicular bisector of $\ol{KL}$ with $\ol{AC}$ and $\ol{BD}$.

\begin{center}
\begin{asy}
  pair A = (-4,0);
  pair B = (0,-3);
  pair C = -A;
  pair D = -B;
  pair K = midpoint(A--B) + dir(90)*(B-A)*0.2;
  pair L = D+K-A;
  pair X = -A+2*foot(K,A,C);
  pair Y = -D+2*foot(L,B,D);
  pair O = origin;
  pair M = midpoint(K--L);
  pair U = midpoint(A--B);
  pair V = midpoint(C--D);

  filldraw(A--B--C--D--cycle, opacity(0.1)+lightred, red);
  draw(A--C, red);
  draw(B--D, red);
  filldraw(A--K--B--cycle, opacity(0.1)+cyan, blue);
  filldraw(C--L--D--cycle, opacity(0.1)+cyan, blue);
  draw(X--Y, deepgreen);
  draw(L--K, deepgreen);
  draw(K--U, black+1.5);
  draw(L--V, black+1.5);
  draw(M--O, black+1.5);

  dot("$A$", A, dir(210));
  dot("$B$", B, dir(340));
  dot("$C$", C, dir(-30));
  dot("$D$", D, dir(160));
  dot("$K$", K, dir(100));
  dot("$L$", L, dir(90));
  dot("$X$", X, dir(305));
  dot("$Y$", Y, dir(200));
  dot("$M$", M, dir(-5));
  dot("$O$", O, dir(225));
  dot("$U$", U, dir(250));
  dot("$V$", V, dir(250));
\end{asy}
\end{center}

\begin{claim*}
  The midpoint of $\ol{XY}$ coincides with the midpoint of $\ol{KL}$.
\end{claim*}
\begin{proof}
  Because
  \begin{align*}
    \ol{XY} &\perp \ol{KL} \parallel \ol{BC} \\
    \ol{MO} &\perp \ol{AB} \\
    \ol{BD} &\perp \ol{AC}
  \end{align*}
  it follows that $\triangle MOY$,
  which was determined by the three lines $\ol{XY}$, $\ol{MO}$, $\ol{BD}$,
  is similar to $\triangle ABC$.
  In particular, it is isosceles with $MY = MO$.
  Analogously, $MX = MO$.
\end{proof}

\begin{remark*}
  It is also possible to simply use coordinates to prove both claims.
\end{remark*}

\paragraph{Second solution (author's solution).}
In this solution, we instead define
$X$ and $Y$ as the intersections of the circles
centered at $K$ and $L$ of equal radii $KA$,
which will be denoted $\omega_K$ and $\omega_L$.
It is clear that $KXLY$ is a rhombus under this construction,
so it suffices to show that $X$ and $Y$ lie on $AC$ and $BD$ (in some order).

\begin{center}
\begin{asy}
  pair A = (-4,0);
  pair B = (0,-3);
  pair C = -A;
  pair D = -B;
  pair K = midpoint(A--B) + dir(90)*(B-A)*0.2;
  pair L = D+K-A;
  pair X = -A+2*foot(K,A,C);
  pair Y = -D+2*foot(L,B,D);

  filldraw(A--B--C--D--cycle, opacity(0.1)+lightred, red);
  draw(A--C, red);
  draw(B--D, red);
  filldraw(A--K--B--cycle, opacity(0.1)+cyan, blue);
  filldraw(C--L--D--cycle, opacity(0.1)+cyan, blue);
  draw(X--L--Y--K--cycle, deepgreen);
  draw(circle(K,abs(K-A)), gray);
  draw(circle(L,abs(L-C)), gray);

  dot("$A$", A, dir(210));
  dot("$B$", B, dir(340));
  dot("$C$", C, dir(-30));
  dot("$D$", D, dir(160));
  dot("$K$", K, dir(250));
  dot("$L$", L, dir(90));
  dot("$X$", X, dir(305));
  dot("$Y$", Y, dir(200));
  label("$\omega_L$", L + abs(L-C)*dir(40), dir(40));
  label("$\omega_K$", K + abs(K-B)*dir(220), dir(220));
\end{asy}
\end{center}

To see this, let $\ol{AC}$ meet $\omega_K$ again at $X'$.
We have
\[ \dang CX'D = \dang BX'C = \dang BX'A = \half \opname{m} \arc{AB} = \opname{m} \arc{CD} \]
where the arcs are directed modulo $360\dg$;
here $\arc{AB}$ is the arc of $\omega_K$ cut out by $\dang AXB$,
and $\arc{DC}$ is the analogous arc of $\omega_L$.
This implies $X'$ lies on $\omega_L$ by the inscribed angle theorem.
Hence $X = X'$, and it follows $X$ lies on $\ol{AC}$.

Analogously $Y$ lies on $BD$.
\begin{remark*}
  The angle calculation above can also be replaced with a length calculation, as follows.

  Let $M$ and $N$ be the projections of $K$ and $L$ onto $\ol{AC}$, respectively.
  Then $X'$ is the reflection of $A$ across $M$;
  analogously, the second intersection $X''$ with $\ol{AC}$
  should be the reflection of $C$ across $N$.
  So to get $X = X' = X''$, we would need to show $AC = 2MN$.

  However, note that $AKLD$ is a parallelogram.
  As $MN$ was the projection of $\ol{KL}$ onto $\ol{AC}$,
  its length should be the same as the projection of $\ol{AD}$ onto $\ol{AC}$,
  which is obviously $\half AC$ because the projection of $D$ onto $\ol{AC}$ is
  exactly the midpoint of $\ol{AC}$ (i.e.\ the center of the rhombus).
\end{remark*}