Rumyantsev V.

The perpendicular line restored at vertex $C$ of parallelogram $A B C D$ to line $C D$ intersects at point $F$ the perpendicular line dropped from vertex $A$ to diagonal $B D$, and the perpendicular line restored from point $B$ to line $A B$ intersects at point $E$ the perpendicular bisector of segment $A C$. In what ratio does side $B C$ divide segment $E F$?
Let point $K$ be symmetric to $A$ with respect to $B$. Then $E$ is the circumcenter of triangle $A C K$. On the other hand, since $B K C D$ is a parallelogram, $A F \perp C K$, so $F$ is the orthocenter of triangle $A C K$ (see the figure). Therefore, the median $CB$ divides $E F$ in the ratio $1: 2$ (see problem $\underline{55595}$).

![](https://cdn.mathpix.com/cropped/2024_05_06_43a9f723b796fae20c45g-15.jpg?height=1126&width=1192&top_left_y=843&top_left_x=427)

Answer

$1: 2$.

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