+311 In triangle $ABC,$ $\angle B = 90^\circ.$ Point $X$ is on $\overline{AC}$ such that $\angle BXA = 90^\circ,$ $BC = 15,$ and $CX = 5$. What is $BX$?
+1926 There's a very simple solution to this. We have
\(AB = BC\)
Also note that Triangle AXB is similar to Triangle ABC
Because of that correlation, we can write
\(AX / AB = AB / AC\\ 5 / AB = AB / 10\\ AB^2 = 50 \\ AB = 5\sqrt 2 = BC\)
Using the other sides of the triangles, we also know that
\(BX / AB = BC / AC\\ BX / AB = AB / AC \\ BX / AB = AB / 10\\ 10BX = AB^2\\ 10 BX = 50\\ BX = 50 / 10 = 5\\\)
Thus, our final answer is 5.
Thanks! :)
