[ [ Law of Cosines $\quad$]

In triangle $ABC$, points $E$ and $F$ are the midpoints of sides $AB$ and $BC$ respectively. Point $G$ lies on segment $EF$ such that $EG: AE = 1: 2$ and $FG = BE$. Find: a) the ratio of the areas of triangles $ABG$ and $AGC$; b) $\angle GCA$, if $\angle AGC = 90^{\circ}$.

#
Let $\angle G C A=\alpha$. Applying the cosine theorem to triangle $A E G$, form an equation in terms of $\sin \boldsymbol{\alpha}$.

## Solution

Let $E G=x$, then $B E=A E=2 x, G F=2 x, A C=2 E F=6 x$. If the distance between the parallel lines $E F$ and $A C$ is $h$, then

$$
S_{\triangle \mathrm{ABG}}=2 S_{\triangle \mathrm{AEG}}=2 \cdot \frac{1}{2} E G \cdot h=x h, S_{\triangle \mathrm{AGC}}=\frac{1}{2} A C \cdot h=3 x h
$$

Therefore, $S_{\triangle \mathrm{ABG}}: S_{\triangle \mathrm{AGC}}=1:3$. Let $\angle G C A=\boldsymbol{\alpha}, A G=t$. Then $\angle E G A=\angle G A C=90^{\circ}-\boldsymbol{\alpha}, t=A C \sin \alpha=6 x \sin \alpha$. Applying the cosine theorem to triangle $A E G$, we get the equation

$$
4 x^{2}=x^{2}+t^{2}-2 x t \cos \left(90^{\circ}-\alpha\right)
$$

or

$$
4 x^{2}=x^{2}+36 x^{2} \sin ^{2} \alpha-12 x^{2} \sin ^{2} \alpha
$$

from which we find that $\sin ^{2} \alpha=\frac{1}{8}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_43a9f723b796fae20c45g-44.jpg?height=508&width=788&top_left_y=1581&top_left_x=635)

## Answer

a) $1: 3 ;$ b) $\arcsin \frac{1}{2 \sqrt{2}}$.

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