\section*{Problem 2 - 261012}

Martin tells his friend Jörg that he has drawn a parallelogram \(A B C D\) in which the perpendicular \(B E\) dropped from \(B\) to the line through \(A\) and \(D\) passes through the intersection point \(S\), where the perpendicular bisector \(s\) of \(A B\) intersects the angle bisector \(w\) of \(\angle B A D\). Jörg claims that the size of the angle \(\angle C B A\) can be determined solely from these statements.

Investigate whether Jörg's claim is true! If this is the case, determine the size of the angle \(\angle C B A\)!

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1512.jpg?height=345&width=604&top_left_y=1178&top_left_x=246)
}

From the given information, it follows:

If we denote the size of the angle \(\angle B A D\) by \(\alpha\), then \(\angle E A S = \angle S A B = \frac{\alpha}{2}\), since \(w\) bisects the angle \(\angle B A D\) and passes through \(S\).

Furthermore, \(A S = S B\) because \(S\) lies on the perpendicular bisector \(s\) of \(A B\). Therefore, the triangle \(A S B\) is isosceles, which implies

\[
\angle S A B = \angle S B A = \frac{\alpha}{2}
\]

(Base angles). Since the perpendicular \(B E\) passes through \(S\), \(A B E\) is a right triangle with \(\angle A E B = 90^{\circ}\) and \(\angle A B E = \frac{\alpha}{2}\). Consequently, by the angle sum property, \(\alpha + \frac{\alpha}{2} = 90^{\circ}\), so \(\alpha = 60^{\circ}\).

Since \(\angle C B A = 180^{\circ} - \alpha\) (adjacent angles in the parallelogram), it follows that \(\angle C B A = 120^{\circ}\).

Thus, Jörg's claim is proven to be true, and the desired angle measure has been determined.