\section*{Exercise 2 - 241012}

In a triangle \(A B C\) with acute interior angles at \(A\) and \(B\), a perpendicular is dropped from \(C\) to \(A B\). Its foot is \(D\). For it, the following holds:

\[
\overline{A C} \cdot \overline{B D}=\overline{B C} \cdot \overline{C D}
\]

Determine the size of the interior angle \(\angle A C B\) from this condition!
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![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1483.jpg?height=328&width=596&top_left_y=1778&top_left_x=730)

Since in triangle \(A B C\) the interior angles at \(A\) and \(B\) are acute, the altitude \(C D\) lies inside the triangle. By assumption, \(\angle C D A = \angle B D C = 90^{\circ}\).

From the assumption, it also follows that \(A C: C D = C B: B C\).

Therefore, the triangles \(A C D\) and \(C B D\) are similar by the similarity theorem "ssw" (side-side-angle), since the respective angles they share are right angles, so the larger of the two mentioned sides lies opposite these angles.

This implies \(\angle D A C = \angle D C B\).

Furthermore, in the right triangle \(A D C\), by the angle sum theorem, \(\angle A C D = 90^{\circ} - \angle D A C\). Due to (1), this gives us

\[
\angle A C B = \angle A C D + \angle D C B = 90^{\circ}
\]