Let $ABC$ be a triangle. Let $X$ be a point on $\overline{AC}$ such that $\angle BXA = 90^\circ.$ If $\angle ABC = 90^\circ,$ $AX = 6,$ and $CX = 6,$ then what is $BX?$

Answer 1

[asy]
unitsize(0.4 cm);

pair A, B, C, X;

A = (0,12);
B = (0,0);
C = (24,0);
X = (6,0);

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

label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$X$", X, S);
label("$6$", (A + X)/2, N);
label("$6$", (C + X)/2, N);
label("$6$", (B + X)/2, S);
[/asy]

Since $\angle ABC = 90^\circ$ and $\angle BXA = 90^\circ,$ quadrilateral $ABXC$ is cyclic. Then
\[\angle CXB = \angle CAB = \angle CAX,\]so $\triangle ABC \sim \triangle BXCa$ by AA. Hence,
\[\frac{AB}{BC} = \frac{BX}{XC} = \frac{6}{6} = 1.\]Therefore, we must have $AB = BC = 6,$ so $BX = \boxed{6}.$