\section*{Problem 3 - 111033}

Given the length of the cathetus \(B C=a\) of a right-angled triangle \(\triangle A B C\) with the right angle at \(C\), for which \(A C: B C=2: 1\) holds.

The angle bisector of the right angle \(\angle A C B\) intersects the circumcircle of the triangle at \(D\) in addition to \(C\).

Calculate the length of the chord \(C D\) as a function of \(a\)!

Hint: According to a well-known theorem of plane geometry, the angle bisector in a triangle divides the opposite side in the ratio of the adjacent sides.
}

According to the problem statement, \(A C=2 a\) and by the Pythagorean theorem (1) \(A B=a \sqrt{5}\). Let \(E\) be the intersection of \(A B\) and \(C D\). Then, according to the theorem mentioned in the hint,

\[
A E: E B=A C: B C=2: 1
\]

From this, due to (1): \(A E=\frac{2}{3} a \sqrt{5}\) and \(E B=\frac{1}{3} a \sqrt{5}\). The line parallel to \(B C\) through \(E\) intersects \(A C\) at \(F\). Then, by one of the Thales' theorems, \(E F: B C=A E: A B=2: 3\), which implies \(E F=\frac{2}{3} a\). (see figure)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1304.jpg?height=457&width=462&top_left_y=1325&top_left_x=797)

Triangle \(\triangle E F C\) is a right triangle with the right angle at \(F\) and, since \(\angle A C E\) is \(45^{\circ}\), it is also isosceles. Therefore, \(E F=F C\). By the Pythagorean theorem, it follows that \(C E=\frac{2}{3} a \sqrt{2}\).

Furthermore, \(\triangle A D E \cong \triangle B C E\); because \(\angle B E C \cong \angle A E D\) (vertical angles) and \(\angle C B A \cong \angle C D A\) (inscribed angles subtending the same arc). Therefore, \(C E: E B=A E: E D\), from which we get

\[
E D=\frac{A E \cdot E B}{C E}=\frac{\frac{2}{3} a \sqrt{5} \cdot \frac{1}{3} a \sqrt{5}}{\frac{2}{3} a \sqrt{2}}=\frac{5}{6} a \sqrt{2}
\]

Thus, we obtain

\[
C D=C E+E D=\frac{2}{3} a \sqrt{2}+\frac{5}{6} a \sqrt{2}=\frac{3}{2} a \sqrt{2}
\]

Adapted from [5]