This is a Rectangle (i didn't name it because it's not important), $ABD$ and $ABC$ are both right triangles, such that: $AB=5, BD=7, AC=9$ The question is to find the length of $MH$?
My Attempt: First, calculate $AD$ and $BC$: $$AD^{2} = BD^{2}-AB^{2} \Leftrightarrow AD = \sqrt{49-25}=\sqrt{24}$$ $$BC^{2}= AC^{2}-AB^{2} \Leftrightarrow BC = \sqrt{81-25}=\sqrt{56}$$
Now let's look at angles : $$\tan{\alpha} = \frac{AD}{AB}=\frac{MH}{BH}=\frac{\sqrt{24}}{5}\Leftrightarrow MH = \frac{BH\sqrt{24}}{5}$$ $$\tan{\theta} = \frac{BC}{AB}=\frac{MH}{AH}=\frac{\sqrt{56}}{5}\Leftrightarrow MH = \frac{AH\sqrt{56}}{5}$$
This leaves us with the following system of equations: $$\begin{cases} MH = \frac{DH\sqrt{56}}{5}\\MH = \frac{AH\sqrt{24}}{5} \end{cases}$$ $$\implies \begin{cases} MH = \frac{AD\sqrt{56}-\sqrt{56}AH}{5}\\MH = \frac{AH\sqrt{24}}{5} \end{cases}$$ $$\implies 5\sqrt{56}-AH\sqrt{56} = AH\sqrt{24}$$ and by some basic arithmetic we got:
$$AH = \frac{5\sqrt{56}}{\sqrt{24}+\sqrt{56}}$$ and now we will plug this result to one of the equations: $$MH = \frac{5\sqrt{56}}{\sqrt{24}+\sqrt{56}} \cdot \sqrt{24} = \frac{7\sqrt{6}-3\sqrt{14}}{2}$$ But i still wondering if this answer is right because it appears to be irrational?
Any Help will be appreciate
I stumbled upon a nice observation, while trying to see for a more straight forward solution.
From triangle similarity, we get
$\frac{HB}{AB} = \frac{MH}{AD}$ and $\frac{AH}{AB} = \frac{MH}{BC}$
adding the two
$\frac{HB}{AB} + \frac{AH}{AB} = \frac{MH}{AD} + \frac{MH}{BC}$ , then
$\frac{1}{MH} = \frac{1}{AD} + \frac{1}{BC} $,
Which is interesting because it says the length of $MH$ is independent $AB$,
Now you can solve for AD and BC with pythagorian theorem and find $MH$
...
$AD = \sqrt{24}$ and $BC=\sqrt{56}$
$\frac{1}{MH} = \frac{1}{\sqrt{24}} + \frac{1}{\sqrt{56}}$
$MH = \frac{\sqrt{24}\sqrt{56}}{\sqrt{56}+\sqrt{24}}$