Parallelogram trigonometry (Sorry for the ambiguous title, couldn't think of a better one)
While leafing through a highschool textbook, I found what looked like an interesting question in trigonometry. My trigonometry skills are borderline 0, but I didn't expect it to be too much of a challenge. Well, I was wrong:
The sides of a parallelogram are $a$ and $b$ and its sharp angle is $\alpha$. The diagnols are $n$ and $m$, and the sharp angle between them is $\beta$.
A. Prove: $\frac{mn}{2ab} = \frac{\sin\alpha}{\sin\beta}$
B. Let: $\alpha = \beta$, $a < b$, $m < n$
Prove: $6a^2 + 2b^2 = 3m^2+n^2$
And in (rough) drawing:

Following the law of cosines (and that $\cos(180-\theta) = -\cos(\theta)$):
$n^2 = a^2 + b^2 - 2ab \cos\alpha$ (in $\Delta ABC$)
$m^2 = a^2 + b^2 - 2ab \cos(180-\alpha) = a^2 + b^2 + 2ab \cos(\alpha)$ (in $\Delta DAC$)
$a^2 = (\frac{m}{2})^2 + (\frac{n}{2})^2 -2 \frac{m}{2} \frac{n}{2} \cos(\beta)$ (in $\Delta AEB$)
$b^2 = (\frac{m}{2})^2 + (\frac{n}{2})^2 -2 \frac{m}{2} \frac{n}{2} \cos(180 - \beta)$ (in $\Delta BEC$)
Expanding the last two equations:
$$a^2 = \frac{m^2}{4} + \frac{n^2}{4} - \frac{mn \cos(\beta)}{2}$$
$$b^2 = \frac{m^2}{4} + \frac{n^2}{4} + \frac{mn \cos(\beta)}{2}$$
$$\Rightarrow$$
$$a^2 + b^2 = \frac{m^2}{2} + \frac{n^2}{2}$$
And that's where I hit a wall. I have six variables, and can't find a way to express them in a fashion which resembles the end result. A major setback is that I couldn't find a way to express both alpha and beta in the same triangle - if I could, then the law of sines will probably be a rescuer.
If possible, I'd like that instead of solving it, maybe you can show me a guideline - where I went wrong, or what I'm missing. Thank you in advance.

For part A, try counting the area of the parallelogram in two different ways, as suggested by Jim Belk.
For part B, notice that your diagram has $n$ and $m$ reversed, since $m<n$. In particular, $\alpha$ should be opposite $m$, not $n$. The modified version of your formula for $m$ is then
$$m^2 = a^2+b^2 -2ab\cos\alpha.$$
Try combining this with your formula
$$4a^2 = m^2 + n^2 -2mn\cos\beta$$
and use the result from part A.
