In triangle $ABC,$ $\angle B = 90^\circ.$ Point $X$ is on $\overline{AC}$ such that $\angle BXA = 90^\circ,$ $AX = 5,$ and $CX = 5$. What is $BX$?
A
5
X
B C
AB = BC
Triangle AXB ≈ Triangle ABC
AX / AB = AB / AC
5 / AB = AB / 10
AB^2 = 50
AB = 5sqrt (2) = BC
BX / AB = BC / AC
BX / AB = AB / AC
BX / AB = AB / 10
10BX = AB^2
10 BX = 50
BX = 50 / 10 = 5
+1234 
