A circle touches two parallel lines and their secant. The segment of the secant, enclosed between the parallel lines, is divided by the point of tangency in the ratio $1: 3$. At what angle does the secant intersect each of the parallel lines?
A segment of the secant, enclosed between parallel lines, is seen from the center of the circle at a right angle.

## Solution

Let the line $l$ intersect the given parallel lines $a$ and $b$ at points $A$ and $B$, respectively, and let the circle with center $O$ touch the lines $a$, $b$, and $l$ at points $C$, $D$, and $M$, respectively. Since $OA$ and $OB$ are the bisectors of angles $CAB$ and $ABD$, and the sum of these angles is $180^{\circ}$, then $\angle AOB = 90^{\circ}$.

Draw the median $OK$ of the right triangle $AOB$. Then $BM = 3AM$, $OK = AK = BK = 2AM$, $KM = AK - AM = AM$. Therefore, the height $OM$ of triangle $AOK$ is also its median. Thus, $AO = OK = AK$, meaning triangle $AOK$ is equilateral. Consequently, $\angle BAO = 60^{\circ}$, $\angle BAC = 2 \angle BAO = 120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-40.jpg?height=388&width=443&top_left_y=1701&top_left_x=813)

## Answer

$60^{\circ}$.