6. In triangle $A B C$, point $D$ lies on side $B C$ and point $E$ lies on side $A C$ such that segments $B D, D E$, and $A E$ are all of equal length. Point $F$ is the intersection of segments $A D$ and $B E$. Angle $C$ is $68^{\circ}$.

What is the measure of angle $F$ in triangle $A F B$?
A) $120^{\circ}$
B) $121^{\circ}$
C) $122^{\circ}$
D) $123^{\circ}$
E) $124^{\circ}$

![](https://cdn.mathpix.com/cropped/2024_04_17_37097645698d56aa28c8g-2.jpg?height=403&width=556&top_left_y=1289&top_left_x=1361)
A6. E) $124^{\circ}$ First, we look at the left side of the picture. We see that $\triangle A E D$ is an isosceles triangle, and thus the angles $\angle E A D$ and $\angle E D A$ are equal. The sum of the angles in a triangle is $180^{\circ}$, so $\angle A E D + 2 \cdot \angle E D A = 180^{\circ}$. Since $\angle A E C$ is a straight angle, it also holds that $\angle A E D + \angle C E D = 180^{\circ}$. From this, it follows that $\angle C E D = 2 \cdot \angle E D A$.

Using the same reasoning on the right side of the picture, starting from the isosceles triangle $\triangle B D E$, we find

![](https://cdn.mathpix.com/cropped/2024_04_17_b71e27e72afbe21d81ceg-3.jpg?height=386&width=536&top_left_y=218&top_left_x=1274)
that $\angle C D E = 2 \cdot \angle D E B$. Due to the angle sum of $180^{\circ}$ in the upper triangle $\triangle E C D$, we see that

$$
\begin{aligned}
180^{\circ} & = \angle E C D + \angle C E D + \angle C D E \\
180^{\circ} & = 68^{\circ} + 2 \cdot \angle E D A + 2 \cdot \angle D E B \\
112^{\circ} & = 2 \cdot (\angle E D A + \angle D E B) \\
56^{\circ} & = \angle E D A + \angle D E B
\end{aligned}
$$

Now, we look at the triangle $\triangle E F D$. The angle sum is $180^{\circ}$ and two of the three angles were already in earlier triangles. Thus, we find that

$$
\begin{aligned}
& 180^{\circ} = \angle D E F + \angle E D F + \angle E F D \\
& 180^{\circ} = \angle D E B + \angle E D A + \angle E F D \\
& 180^{\circ} = 56^{\circ} + \angle E F D \\
& 124^{\circ} = \angle E F D
\end{aligned}
$$

Finally, we note that $\angle A F B$ and $\angle E F D$ are vertical angles and thus equal. Therefore, the angle at the question mark is $124^{\circ}$.