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 $BX$?

MeIdHunter Jul 24, 2024

#1**+1 **

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! :)

NotThatSmart Jul 24, 2024