7. Given a rectangular trapezoid $\mathrm{ABCE}$, the bases of which $\mathrm{BC}$ and $\mathrm{AE}$ are equal to 5 and 7, respectively. The smaller lateral side $\mathrm{AB}$ is equal to $\mathrm{BC}$. On $\mathrm{AB}$, a point $\mathrm{F}$ is marked such that $\mathrm{BF}: \mathrm{FA}=2: 3$, on $\mathrm{AC}$, a point $\mathrm{G}$ is marked such that $\mathrm{AG}: \mathrm{GC}=4: 1$; on $\mathrm{AE}$, a point $\mathrm{D}$ is marked such that $\mathrm{AD}: \mathrm{DE}=5: 2$. Determine the degree measure of angle DFG.
Solution. Construct perpendiculars GI and GH.

![](https://cdn.mathpix.com/cropped/2024_05_06_9fc9f23f10a928994848g-03.jpg?height=468&width=628&top_left_y=751&top_left_x=771)

1) $\Delta G I D = \Delta G F H$ - by two legs, since $G I = G H = 4; F H = I D = 1$, therefore $F G = G D$, $\angle F G H = \angle D G I = \alpha$.
2) $G I A H$ is a square, so $\angle I G H = 90^{\circ}, \angle I G H = \angle F G H + \angle I G F = \alpha + \angle I G F$.
3) $\angle D G F = \angle D G I + \angle I G F = \alpha + \angle I G F = 90^{\circ}$.
4) $\triangle D F G$ is a right isosceles triangle, since $F G = G D$, and thus $\angle D F G = 45^{\circ}$.

Answer: 45.