## Task A-2.4.

All vertices of the hexagon $A B C D E F$ lie on a circle with diameter $\overline{A D}$. The line $B F$ intersects the lines $A D$ and $C E$ at points $G$ and $H$ respectively. If $\varangle F E H=56^{\circ}, \varangle D G B=124^{\circ}$ and $\varangle D E C=34^{\circ}$, determine $\varangle C E B$.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_afa6864dec9cf901fc06g-11.jpg?height=526&width=699&top_left_y=1593&top_left_x=610)

From the cyclic nature of quadrilateral $B C E F$, it follows that

$$
\varangle F B C=180^{\circ}-\varangle F E C=180^{\circ}-\left(180^{\circ}-\varangle F E H\right)=56^{\circ}
$$

Now we see that $\varangle B G D+\varangle C B G=180^{\circ}$, so we conclude that $G D \| B C$, or $A D \| B C$.

Therefore, with the equality of angles subtended by the chord $\overline{C D}$, we have

$$
\varangle C D A=90^{\circ}-\varangle D A C=90^{\circ}-\varangle D E C=56^{\circ}
$$

Consider triangle $B C D$. The angle $\varangle D B C$ is equal to the angle $\varangle D E C$ (inscribed angles subtended by the same chord) and measures $34^{\circ}$. The angle $\varangle B C D$ is supplementary to the angle $\varangle C D A$ (since lines $A D$ and $B C$ are parallel), so it measures $124^{\circ}$. The remaining angle of this triangle is

$$
\varangle C D B=180^{\circ}-\varangle D B C-\varangle B C D=180^{\circ}-34^{\circ}-124^{\circ}=22^{\circ}
$$

Finally, the angles $\varangle C D B$ and $\varangle C E B$ are inscribed angles subtended by the same chord, so they are equal. Therefore, the measure of the desired angle $\varangle C E B$ is $22^{\circ}$.

Note: Scoring schemes for solutions similar to the one presented follow this structure: 4 points for proving the parallelism of lines $A D$ and $B C$; 3 points for finding the measure of angle $\varangle C D A$ (or proving an equivalent statement, such as the parallelism of lines $C D$ and $B F$), with 1 point given for using Thales' theorem over the diameter $\overline{A D}$; 3 points for completing the solution.