The Law of Sines

In triangle $ABC$, it is known that $AB = BC, \angle BAC = 45^{\circ}$. Line $MN$ intersects side $AC$ at point $M$ and side $BC$ at point $N$, with $AM = 2 \cdot MC, \angle NMC = 60^{\circ}$. Find the ratio of the area of triangle $MNC$ to the area of quadrilateral $ABNM$.
Let $M C=a$ and, applying the Law of Sines, express the segment $C N$ from triangle $C M N$ in terms of $a$.

## Solution

Let $M C=a$. Then $A C=3 a, B C=\frac{3 a}{\sqrt{2}}$. By the Law of Sines in triangle $M C N$, we find that

$$
C N=M C \cdot \frac{\sin \angle N M C}{\sin \angle C N M}=a \cdot \frac{\sin 60^{\circ}}{\sin 75^{\circ}}=\frac{\frac{a \sqrt{3}}{2}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{2 a \sqrt{3}}{\sqrt{6}+\sqrt{2}}
$$

Then

$$
\begin{gathered}
S_{\triangle \mathrm{MNC}}=\frac{1}{2} \cdot C M \cdot C N \cdot \sin 45^{\circ}=\frac{1}{2} \cdot a \cdot \frac{2 a \sqrt{3}}{\sqrt{6}+\sqrt{2}} \cdot \frac{\sqrt{2}}{2}=\frac{a^{2}(3-\sqrt{3})}{4}, \\
S_{\mathrm{ABNM}}=S_{\triangle \mathrm{ABC}}-S_{\triangle \mathrm{MNC}}=\frac{9 a^{2}}{4}-\frac{a^{2}(3-\sqrt{3})}{4}=\frac{a^{2}(6+\sqrt{3})}{4}
\end{gathered}
$$

Therefore,

$$
\frac{S_{\triangle M N C}}{S_{A B N M}}=\frac{a^{2}(3-\sqrt{3})}{4}: \frac{a^{2}(6+\sqrt{3})}{4}=\frac{3-\sqrt{3}}{6+\sqrt{3}}=\frac{7-3 \sqrt{3}}{11} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-30.jpg?height=378&width=640&top_left_y=2335&top_left_x=709)

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