Geometry

Question

0

22

1

+478

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

NotThatSmart · Answer 1 · 2024-07-24T10:22:42

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! :)