1. ## Similarity Problem

Hi, in this problem, I'm not sure what the diagram looks like for a start, also I'm not sure how to solve it. I think congruency/similarity is involved.

Let XYZ be a triangle with a right angle at Z. The midpoint of hypotenuse XY is twice as far from side YZ as it is from side XZ. Find the length of side XZ, if XY = 10mm

2. Hello, BG5965!

If you sketch the diagram, it's quite simple . . .

Let $\displaystyle XYZ$ be a triangle with a right angle at $\displaystyle Z.$
The midpoint $\displaystyle M$ of hypotenuse $\displaystyle XY$ is twice as far from side $\displaystyle YZ$ as it is from side $\displaystyle XZ.$
Find the length of side $\displaystyle XZ$, if $\displaystyle XY = 10$ mm
Let $\displaystyle d$ = distance from $\displaystyle M$ to $\displaystyle XZ.$
Code:
    X *
|\
| \
2d |  \ 5
|   \
|  d \  M
* - - *
|\    :\
| \   : \
2d |  \  :  \ 5
|   \ :2d \
|    \:    \
Z *-----*-----* Y
d     d
The diagram is filled with congruent right triangles!

Using Pythagorus on $\displaystyle \Delta XYZ\!:\;\;(2d)^2 + (4d)^2 \:=\:10^2 \quad\Rightarrow\quad 4d^2 + 16d^2 \:=\:100$

. . $\displaystyle 20d^2 \:=\:100 \quad\Rightarrow\quad d^2 \:=\:5 \quad\Rightarrow\quad d \:=\:\sqrt{5}$

Therefore: .$\displaystyle XZ \:=\:4d \;=\;4\sqrt{5}$

3. But with XY = 10, that diagram isn't correct, because although you put d and 2d, when I draw that diagram out with a ruler, it doesn't work!!!

Is there a special way of doing it or do I assume 'not to scale'?

Thanks