\section*{Problem 4 - 061044}

Given is a triangle \(\triangle A B C\); as usual, let \(B C=a, C A=b\) and \(\gamma\) be the degree measure of the angle \(\angle A C B\).

Construct a square whose area is \(2 a b \cdot|\cos \gamma|\)!
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1. Construct the square over \(|A B|=c\).

2. Complete the side \(B B^{*}\) of the square with the triangle \(B^{\prime} B B^{*}\) congruent to triangle \(A B C\).

3. Form the parallelogram \(B^{\prime} C^{\prime} C B\) from the sides \(\left|B B^{\prime}\right|\) and \(|B C|\); this has the area \(A_{P}=a \cdot h=a b \sin \left(90^{\circ}-\gamma\right)=a b \cos (\gamma)\).

4. Form a rectangle \(P Q C B\) of equal area from the parallelogram; double the rectangle to form rectangle \(R S C B\); this has the area \(A=2 h \cdot a\).

5. Extend the rectangle side \(|C S|\) by \(a=|R S|\), with the endpoint being \(T\).

6. Construct the Thales circle over the segment \(|C T|\) (the center being \(M\)).

7. Determine the intersection of the line through \(|R S|\) with the Thales circle; the intersection point being \(X\).

8. Then, for the right triangle \(C T X\), according to the altitude theorem: The rectangle formed by the segments of the altitude ( \(2 h\) and \(a\) ) to the hypotenuse is as large as the square over the altitude; in other words: \(A=2 h \cdot a=2 b \cos (\gamma) \cdot a=2 a b \cos (\gamma)\).

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1244.jpg?height=1331&width=1108&top_left_y=588&top_left_x=474)