4B. Given four points $A, B, C$ and $D$ in the plane such that $\overline{A B}=\overline{A C}$ and $\overline{A D}=\overline{B D}$. Let $E$ be a point on the line $A C$, such that $A$ is between $E$ and $C$. If the angles $\alpha=\measuredangle B A E$ and $\beta=\measuredangle A D C$ satisfy $\alpha+\beta=200^{\circ}$, determine the angle $\varphi=\measuredangle B C D$.
Solution. Since triangle $ABC$ is isosceles with base $BC$, and $\alpha = \measuredangle BAE$, it follows that

$$
\measuredangle ACB = \measuredangle CBA = \frac{180^{\circ} - \measuredangle BAC}{2} = \frac{\alpha}{2}
$$

Since triangle $ABD$ is also isosceles with base $AB$, we have

$$
\measuredangle DBA = \measuredangle BAD = \frac{180^{\circ} - \beta}{2}.
$$

Therefore, it follows that

$$
\begin{aligned}
\varphi & = \angle CBD = \measuredangle CBA - \measuredangle DBA = \frac{\alpha}{2} - \frac{180^{\circ} - \beta}{2} = \\
& = \frac{\alpha + \beta}{2} - 90^{\circ} = 10^{\circ}
\end{aligned}
$$

## Third Year

![](https://cdn.mathpix.com/cropped/2024_06_05_ce8ee69309f5a5d5bef7g-07.jpg?height=358&width=486&top_left_y=249&top_left_x=978)