5. In a right triangle $ABC$ ( $\angle C$ - right), a point $D$ is taken on side $BC$ such that $\angle CAD=30^{\circ}$. From point $D$, a perpendicular $DE$ is dropped to $AB$. Find the distance between the midpoints of segments $CE$ and $AD$, given that $AC=3 \sqrt{3}, DB=4$.
Answer: $\frac{9 \sqrt{57}}{38}$

## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_9c68f41690a65bb4cc11g-3.jpg?height=850&width=1004&top_left_y=601&top_left_x=209)

Let point $G$ be the midpoint of segment $A D$, and point $F$ be the midpoint of segment $C E$. We need to find the length of segment $G F$.

Since $\angle A C B = \angle D E A = 90^{\circ}$, a circle can be circumscribed around quadrilateral $A C D E$, and $G$ is the center of this circle (since $A D$ is the diameter). Therefore, $CG = GE = GD$ (as radii of the circle), meaning $\triangle C G D$ and $\triangle C G E$ are isosceles, and thus $G F$ is the altitude in $\triangle C G E$.

Since $\angle C A D = 30^{\circ}$ is an inscribed angle subtending arc $C D$, and $\angle C G D$ is a central angle subtending the same arc, $\angle C G D = 60^{\circ}$, making $\triangle C G D$ equilateral.

Let $\angle E A D = \alpha$, then $\angle E C D = \alpha$ as an inscribed angle subtending the same arc $D E$. Therefore, $\angle G C F = 60^{\circ} - \alpha$, $\angle C A B = 30^{\circ} + \alpha$, and thus $\angle C B A = 90^{\circ} - (30^{\circ} + \alpha) = 60^{\circ} - \alpha$. This means $\triangle C G F$ and $\triangle A B C$ are similar by two angles ($\angle C B A = \angle G C F = 60^{\circ} - \alpha$, $\angle A C B = \angle G F C = 90^{\circ}$).

From the similarity of $\triangle C G F$ and $\triangle B A C$, we have $\frac{C G}{A B} = \frac{G F}{A C} \Rightarrow G F = \frac{C G \cdot A C}{A B} = \frac{C D \cdot A C}{A B}$. From $\triangle C A D$, we get $A D = 2 C D$ (since $\angle C A D = 30^{\circ}$) and by the Pythagorean theorem, $C D = 3$. From $\triangle A B C$, by the Pythagorean theorem, $A B = 2 \sqrt{19} \Rightarrow G F = \frac{3 \cdot 3 \sqrt{3}}{2 \sqrt{19}} = \frac{9 \sqrt{57}}{38}$.

## Recommendations for checking.

It is noted that a circle can be circumscribed around quadrilateral $A C D E$, and $G$ is the center of this circle - 1 point.

If, in addition to the above, the sides $A B$ and $C D$ are calculated, then +1 point.