Problem: Let $ABC$ be a triangle with $\angle BCA = 90\dg$,
and let $D$ be the foot of the altitude from $C$.
Let $X$ be a point in the interior of the segment $CD$.
Let $K$ be the point on the segment $AX$ such that $BK = BC$.
Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$.
Let $M = \ol{AL} \cap \ol{BK}$.
Prove that $MK = ML$.

Solution: Let $\omega_A$ and $\omega_B$ be the circles through $C$
centered at $A$ and $B$;
extend rays $AK$ and $BL$ to hit $\omega_B$ and $\omega_A$ again at $K^\ast$, $L^\ast$.
By radical center $X$,
we have $KLK^{\ast}L^{\ast}$ is cyclic,
say with circumcircle $\omega$.

\begin{center}
  \begin{asy}
    size(10cm);
    pair A = Drawing("A", dir(180), dir(225));
    pair B = Drawing("B", dir(0), dir(-45));
    pair C = Drawing("C", dir(140), 1.2*dir(110));
    draw(A--B--C--cycle);
    pair D = Drawing("D", (C.x,0), dir(-90));
    pair X = Drawing("X", 0.55*C+0.45*D, dir(-80));

    pair K = Drawing("K", IP(A--X, Drawing(CP(B,C))), dir(-50));
    pair L = Drawing("L", IP(B--X, Drawing(CP(A,C))), dir(240));
    pair K1= Drawing("K^\ast", 2*foot(B,A,K)-K, dir(60));
    pair L1= Drawing("L^\ast", 2*foot(A,B,L)-L, dir(L-B));
    draw(A--K1);
    draw(B--L1);
    draw(circumcircle(K,L,K1), dashed);

    /*
    pair O = Drawing("O", circumcenter(K,L,K1), dir(90));
    draw(A--O--B, dotted);
    draw(O--D, dotted);
    */

    pair M = Drawing("M", extension(A,L,B,K), dir(270));
    draw(A--L, dotted);
    draw(B--K, dotted);

    clip((1.2,-0.4)--(1.2,2.3)--(-2.4,2.3)--(-2.4,-0.4)--cycle);
  \end{asy}
\end{center}

By orthogonality of $(A)$ and $(B)$ we find that
$\ol{AL}$, $\ol{AL^\ast}$,
$\ol{BK}$, $\ol{BK^\ast}$ are tangents to $\omega$
(in particular, $KLK^{\ast}L^{\ast}$ is harmonic).
In particular $\ol{MK}$ and $\ol{ML}$ are tangents to $\omega$,
so $MK = ML$.