Let ABC be a triangle with right angle C. The mid-point of the hypotenuse AB is twice as far from the side BC as it is from side AC. Find the length of the side AC, if AB = 10cm.
The wording of your question is a bit confusing. By the distance from midpoint of $\displaystyle AB $ to $\displaystyle BC $ and $\displaystyle AC$, do you mean the distance $\displaystyle DE $ and $\displaystyle DF $ as I have in my diagram?
This is how I did the calculation, if this is wrong, just let me know.
Answer
Let $\displaystyle DE=x, DF=2x$
As point, by letting $\displaystyle \angle{B}=\theta$ and $\displaystyle \angle{A}=90-\theta$, we see that the triangles $\displaystyle AED $and $\displaystyle DFB $are the same. Thus
$\displaystyle AE=EC=2x$, $\displaystyle CF=FB=x$
Calculate for x with Pythagoras' theorem, and you'll find AC.
If my assumption is wrong, just let me know.
The 'Pythagoras' reference:
The distance BC is 2x and the disance CA is 4x
$\displaystyle (2x)^2 + (4x)^2 = 10^2 $
$\displaystyle 4x^2 + 16x^2 = 10^2 $
$\displaystyle x^2 (4+16) = 10^2 $
$\displaystyle x^2 = \frac {10^2}{(4+16)} $
$\displaystyle x = \sqrt { \frac {10^2}{(4+16)} } $
NOTE:
This is being done ( not specificly or exactly for your benefit ) so that I can practice using Latex.
I'm sorry that wasn't clear.
My apologies.
Obviously my attempt at clarity wasn't.
You probably used this to answer the question anyway:
$\displaystyle \bar{AC} = 10 \times \sin \left ( \tan^{-1} \left ( \frac{2x}{x} \right ) \right ) $
Again, sorry for the obfuscation.
I can use trig because I had a very good 6th grade teacher. The teacher was adamant that we know what a trig function did -- not merely able to write that tangent = sine/cosine or that sine = opposite / hypotenuse.
I can use trig because I practiced solving and creating trig problems.
There are many reasons that I can use trig, but those are the primaries.