<s>[INST] The area of $\triangle AMN$ where $MN$ is midsegment $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. If the area of $\triangle ABC$ is $24$ $cm^2,$ what is the area of $\triangle AMN$?

We can say for sure that $MN\parallel BC$ and $MN=\dfrac12 BC$. I don't know what to do next. Any help would be appreciated. Thank you in advance!
 [/INST] We know that $$S_\triangle = \frac{1}{2}ab\sin\alpha$$
where $a$ and $b$ are sides and $\alpha$ the angle between them. Using this, we have
$$\frac{S_{ABC}}{S_{AMN}} = \frac{\frac{1}{2}\cdot \color{red}{AB}\cdot \color{blue}{AC}\cdot \sin\alpha}{\frac{1}{2}\cdot \color{red}{AM}\cdot \color{blue}{AN} \cdot \sin \alpha} = \color{red}{2}\cdot \color{blue}{2}=4 
 \Rightarrow S_{AMN} = \frac{S_{ABC}}{4} = \frac{24}{4} = 6$$
after some cancellations.
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