In the right-angled triangle $ABC$, the hypotenuse is $AB$, and the projection of vertex $C$ onto the hypotenuse is $D$. On the line $AB$, there are points $E$ and $F$ such that the magnitude and direction of $ED$ and $DF$ are equal to $AB$. The line through $E$ parallel to $AC$ and the line through $F$ parallel to $BC$ intersect at point $G$. Express the distance $CG$ in terms of the sides of the triangle.
Triangle $E F G$ is similar to triangle $A B C$, with sides twice as long and parallel to the corresponding sides of triangle $A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d862584bc09c9411ed91g-1.jpg?height=339&width=566&top_left_y=259&top_left_x=758)

The line $G C$ intersects $A B$ at point $H$, and $A C$ and $B C$ are the midlines of triangles $E G H$ and $F G H$, respectively, because they are parallel to $E G$ and $F G$ and half their length. Thus,

$$
G C = C H, \quad A H = E A = E D - A D = A B - A D = D B
$$

and similarly, $B H = D A$. From the right triangle $C D H$, by the Pythagorean theorem,

$$
G C^{2} = C H^{2} = C D^{2} + D H^{2} = C D^{2} + (A H - A D)^{2} = C D^{2} + (D B - A D)^{2}
$$

Denoting the sides $A B$, $B C$, and $C A$ by $c$, $a$, and $b$ respectively, we can express the double area of triangle $A B C$ in two ways:

$$
c \cdot C D = a b, \quad \text { hence } \quad C D = a b / c
$$

To determine $A D$, note that the circle with diameter $B C$ passes through $D$ and is tangent to $A C$. Thus, for the tangent $A C$ and the secant $A B$ from $A$,

$$
\begin{gathered}
b^{2} = A C^{2} = A D \cdot A B = A D \cdot c \\
\text { hence } \quad A D = b^{2} / c
\end{gathered}
$$

and similarly,

$$
D B = a^{2} / c
$$

Using these,

$$
G C^{2} = \frac{a^{2} b^{2}}{c^{2}} + \left(\frac{a^{2}}{c} - \frac{b^{2}}{c}\right)^{2} = \frac{a^{4} - a^{2} b^{2} + b^{4}}{c^{2}}
$$

which is the desired expression. Here, one side can be easily eliminated using the other two based on the Pythagorean theorem.

Fazekas Tibor (Győr, Czuczor G. Bencés Gimn., II. o. t.)

Szepesi László (Sopron, Széchenyi I. Gimn., II. o. t.)