7. Given a rectangular trapezoid $\mathrm{ABCE}$, the bases of which $\mathrm{BC}$ and $\mathrm{AE}$ are equal to 3 and 4, respectively. The smaller lateral side $\mathrm{AB}$ is equal to $\mathrm{BC}$. On $\mathrm{AE}$, a point $\mathrm{D}$ is marked such that $\mathrm{AD}: \mathrm{DE}=3: 1$; on $\mathrm{AD}$, a point $\mathrm{F}$ is marked such that $\mathrm{AF}: \mathrm{FD}=2: 1$; on $\mathrm{BD}$, a point $\mathrm{G}$ is marked such that $\mathrm{BG}: \mathrm{GD}=1: 2$. Determine the degree measure of angle $\mathrm{CFG}$.
Solution. Construct the height $I H$ such that $G \in I H$ and connect points C and G.

![](https://cdn.mathpix.com/cropped/2024_05_06_b77a4ddf94595f81d054g-03.jpg?height=525&width=694&top_left_y=1331&top_left_x=201)

1) $\triangle I G C = \triangle G F H$ - by two legs, since $I C = G H = 2, I G = H F = 1$, therefore $F G = G C$, $\angle I C G = \angle F G H = \alpha$, and $\angle H F G = \angle I G C = 90^{\circ} - \alpha$.
2) $\triangle F G C$ is a right isosceles triangle, since $F G = F D$,

$$
\begin{aligned}
& \angle H G F + \angle F G C + \angle I G C = 180^{\circ} \\
& \angle F G C = 180^{\circ} - \angle H G F - \angle I G C \\
& \angle F G C = 180^{\circ} - \alpha - 90^{\circ} + \alpha = 90^{\circ}
\end{aligned}
$$

Thus, $\angle C F G = 45^{\circ}$.

Answer: 45.