\section*{Problem 3 - 091023}

Given are two line segments of lengths \(m\) and \(n\) (with \(n < m\)).

a) Perform the following construction:

Around an arbitrary point \(Y\) on a line \(g\), draw a circle \(k_{1}\) with radius \(m\). One of the intersection points of \(g\) and \(k_{1}\) is called \(A\), the other \(E\).

From \(A\), measure the segment \(A B\) with \(A B = n\) on \(g\) such that \(B\) lies between \(A\) and \(Y\) (this is possible because \(n < m\)). From \(B\), measure the segment \(B C\) with \(B C = m\) on \(g\) such that \(A\) lies between \(B\) and \(C\) (this is again possible because \(n < m\)). Draw a circle \(k_{2}\) with radius \(B C\) around \(C\). One of the intersection points of \(k_{1}\) and \(k_{2}\) is called \(D\).

b) Determine the length \(x\) of the segment \(A D\)!
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a) If one proceeds as indicated, the following image is created:

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1273.jpg?height=503&width=1031&top_left_y=831&top_left_x=501)

b) From the construction, it follows that \(A B=n \quad ; \quad A Y=D Y=E Y=m\). Furthermore, \(A D=x=B D\).

According to a theorem of elementary geometry, the perpendicular dropped from \(D\) to \(g\) bisects the segment \(A B\). Let its foot be \(F\). By the Thales' theorem, the triangle \(\triangle A E D\) is right-angled. In it, by the theorem of Euclid (cathetus theorem):

\[
x^{2}=\frac{n}{2}(2 m)=m \cdot n \quad \text { thus } \quad x=\sqrt{m \cdot n}
\]