Problem 5. In a convex quadrilateral $A B C D$ (trapezoid), points $E, F, G$ are the midpoints of sides $A D, D C$ and $A B$ respectively. Given that $G E \perp A D$ and $G F \perp C D$. Calculate the angle $\measuredangle A C B$.
Solution. From the condition of the problem $\overline{A E}=\overline{E D}, \measuredangle G E A=\measuredangle D E G=90^{\circ}$ and $G E$ is a common side, so according to SAS (Side-Angle-Side) the triangles $A G E$ and $D G E$ are congruent. From the congruence, it follows that $\overline{A G}=\overline{G D}$. Similarly, $\triangle G F D \cong \triangle G F C$ from which $\overline{G D}=\overline{G C}$.

Thus $\overline{A G}=\overline{G D}=\overline{G C}$ and $\overline{A G}=\overline{G B}$ from the condition that point $G$ is the midpoint of side $A B$, which means $\overline{G C}=\overline{G B}$, or that $\triangle G B C$ is isosceles.

Let us denote $\measuredangle C A G=\alpha$. Then $\measuredangle G C A=\alpha$, so

![](https://cdn.mathpix.com/cropped/2024_06_05_a1a5b9bf3a5d296219b1g-10.jpg?height=433&width=781&top_left_y=634&top_left_x=720)

$$
\measuredangle C G B=\measuredangle C A G+\measuredangle G C A=2 \alpha
$$

Furthermore,

$$
\measuredangle B C G=\measuredangle G B C=\frac{180^{\circ}-\measuredangle C G B}{2}=90^{\circ}-\alpha .
$$

Finally,

$$
\measuredangle A C B=\measuredangle G C A+\measuredangle B C G=\alpha+90^{\circ}-\alpha=90^{\circ}
$$

## VIII Department