# Task No. 8.1

## Condition:

Given triangle $\mathrm{ABC}$, where $2 \mathrm{BC}=\mathrm{AC}$ and angle $\mathrm{C}=74^{\circ}$. On ray $\mathrm{BC}$, segment $\mathrm{CD}=\mathrm{CB}$ is laid out. Then, from point $\mathrm{D}$, a perpendicular is drawn to the line containing the median of triangle $\mathrm{ABC}$, drawn from vertex $\mathrm{B}$, and the intersection point is $\mathrm{F}$. What is the measure of angle CDF? Express your answer in degrees.
Answer: 37

## Exact match of the answer -1 point

## Solution.

Let $\mathrm{F}^{\prime}$ be the midpoint of segment $\mathrm{AC}$. Then $\mathrm{F}^{\prime} \mathrm{C}=0.5 \mathrm{AC}=\mathrm{BC}=\mathrm{CD}=0.5 \mathrm{BD}$. From this, it follows that triangle BDF' is a right triangle, since its median is equal to half the side to which it is drawn. Therefore, the angle $\mathrm{DF}^{\prime} \mathrm{B}=90^{\circ}$, and $\mathrm{DF}^{\prime}$ is perpendicular to the median of angle B. This means that point $\mathrm{F}^{\prime}$ coincides with F. Note that triangle FCD is isosceles, since segments FC and CD are equal, and therefore the angles CFD and CDF, as the angles at the base, are equal. Angle C is the exterior angle for triangle FCD, so $\angle \mathrm{C}=\angle \mathrm{CFD}+\angle \mathrm{CDF}=2 \angle \mathrm{CDF}$. Therefore, $\angle \mathrm{CDF}=0.5 \angle \mathrm{C}=37^{\circ}$.