In the trapezoid $A B C D$, the length of the leg $B C$ is equal to the sum of the lengths of the bases $A B$ and $C D$. The midpoint of the leg $D A$ is $F$. What is the measure of angle $B F C$?
I. Solution. We will show that the angle $BFC$ is a right angle.

# $1987-01-027-3 . eps$ 

1. Figure

Let the midpoint of the side $BC$ be $O$. Then $OF$ is the midline of the trapezoid, and twice its length is equal to the sum of the lengths of the bases, which, according to the condition, is equal to the length of the side $BC$:

$$
2 OF = AB + CD = BC = BO + OC
$$

Since $O$ is the midpoint, $BO = OC$, so due to (1), $OF = OB = OC$, and thus point $F$ lies on the Thales circle of segment $BC$. This indeed means that the angle $BFC$ is a right angle.

II. Solution. Reflect the trapezoid over the midpoint $F$ of the side $AD$. Due to the parallelism of the bases $DC$ and $AB$, the image of point $B$, denoted as $B'$, lies on the line $DC$, and similarly, the image of point $C$, denoted as $C'$, lies on the line $AB$ (2nd figure). Due to the reflection, the quadrilateral $B'CB'C'$ is a parallelogram. Moreover, since $C'B = C'A + AB = CD + AB$, the condition implies that $C'B = BC$, meaning the quadrilateral is a rhombus. The diagonals $CC'$ and $BB'$ are therefore perpendicular, and thus the angle $BFC$ is a right angle.

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2. Figure