In the right-angled triangle $A B C$, the midpoint of the leg $B C$ is $D$, a point on the line $A D$ is $E$, the line $C E$ intersects the hypotenuse $A B$ at $F$, and the perpendicular from $F$ to $B C$ intersects $B E$ at $G$. What is the geometric locus of the point $G$ as $E$ runs along the line $A D$?
Let's choose the axes of our coordinate system to be the lines $CB$ and $CA$, and let the coordinates be $A(0, a)$, $B(b, 0)$, so $D(b / 2, 0)$. Any point $E$ on the line $AD$ can be written as $E(\lambda b / 2, (1-\lambda) a)$, where $\lambda$ is any real number. The equation of the line $CE$ is

$$
y = \frac{2(1-\lambda) a}{\lambda b} x
$$

provided $\lambda \neq 0$, i.e., if $E$ is not the same as $A$; we assume this for now. On the other hand, the equation of the line $AB$ is

$$
x / b + y / a = 1
$$

and thus the abscissa of the intersection point $F$ is

$$
x_{0} = \lambda b / (2 - \lambda)
$$

if $\lambda \neq 2$. If $\lambda = 2$, then $E$ is at the point $(b, -a)$, the reflection of $A$ over $D$, $CE$ is parallel to $AB$, $F$ does not exist, and neither does $G$.

The equation of the line $BE$ is

$$
y = \frac{2(1-\lambda) a}{b(\lambda - 2)} (x - b)
$$

and according to the construction, $G$ is the point on this line with abscissa $x_{0}$, so its ordinate is

$$
y_{0} = 4 a \left( \frac{1 - \lambda}{2 - \lambda} \right)^2
$$

and from (1),

$$
\lambda = 2 x_{0} / (x_{0} + b) \quad (x_{0} \neq -b)
$$

eliminating $\lambda$, we get the following relationship between the coordinates of $G$:

$$
y_{0} = \frac{a}{b^2} (x_{0} - b)^2
$$

This relationship is also valid in the excluded case $\lambda = 0$, since then $E$ is outside, and $F$ and $G$ both coincide with $A$, and (2) holds with the coordinates of $A$. Thus, we have found that, except for the single point $(b, -a)$, to every point on the line $AD$ there corresponds a point $G$ constructed as required, and for the coordinates of $G$, (2) always holds. This relationship is the equation of a parabola, whose vertex is the point $B$, the tangent at the vertex is the line $BC$, and one point of the parabola is $A$. (It is easy to show that the focus of the parabola is the intersection of the perpendiculars raised at $B$ to $BC$ and at $D$ to $DA$.)

![](https://cdn.mathpix.com/cropped/2024_05_02_82a7aff01b9b3580cfd3g-1.jpg?height=719&width=1222&top_left_y=1611&top_left_x=446)

Conversely, we need to check which points of the parabola (2) belong to the desired geometric locus. Let $G(x_{0}, y_{0})$ be a point for which (2) holds. This can only come from the $\lambda$ value given by (1'), and from $E$ we indeed get back $G$ as described above, provided $x_{0} \neq -b$, since the value of (1') cannot be 2. Therefore, only the point $G^{*}(-b, 4a)$ on the parabola with abscissa $x_{0} = -b$ does not belong to the desired geometric locus, and the geometric locus is obtained from the parabola (2) by removing the point $G^{*}$.

Remark. The problem can also be solved using the theorems of projective geometry.[^0]

[^0]:    ${ }^{1}$ See, for example: Vigassy Lajos: Geometric Transformations, Középiskolai Szakköri Füzetek, Tankönyvkiadó, Budapest, 1963.