Kite $A B C D$ is symmetric with respect to the diagonal $A C$. The length of $A C$ is $12 \mathrm{~cm}$, the length of $B C$ is $6 \mathrm{~cm}$, and the internal angle at vertex $B$ is a right angle. Points $E$, $F$ are given on sides $A B, A D$ such that triangle $E C F$ is equilateral.

Determine the length of segment $E F$.

(K. Pazourek)
Triangles $A C B$ and $A C D$ are symmetric with respect to the common side $A C$, points $E, F$ lie on sides $A B, A D$ and triangle $E C F$ is equilateral, so this triangle is also symmetric with respect to $A C$. In particular, segments $E F$ and $A C$ are perpendicular; their intersection is denoted by $G$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5c762d128fb89da6abfdg-3.jpg?height=591&width=671&top_left_y=710&top_left_x=698)

The internal angles of an equilateral triangle are $60^{\circ}$. The axis of symmetry, or the height of the equilateral triangle, bisects this angle and divides the triangle into two congruent right triangles with the remaining internal angles of $30^{\circ}$ and $60^{\circ}$. The ratio of the hypotenuse to the shorter leg in these smaller triangles is exactly $2: 1$. We will use these general observations in our problem in several ways:

First, the line $A C$ is the axis of symmetry of triangle $E C F$, so angle $A C E$ has a measure of $30^{\circ}$. Second, the ratio of the hypotenuse to the shorter leg in the right triangle $A B C$ is given as $2: 1$, so it is half of an equilateral triangle, with the remaining internal angles $B A C$ and $A C B$ having measures of $30^{\circ}$ and $60^{\circ}$.

From this, we deduce that angle $B C E$ has a measure of $30^{\circ}$ (the difference between angles $A C B$ and $A C E$). Therefore, triangle $A B C$ consists of three mutually congruent right triangles $A G E$, $C G E$, and $C B E$ (they are congruent in their internal angles and each pair has a common side).

![](https://cdn.mathpix.com/cropped/2024_04_17_5c762d128fb89da6abfdg-3.jpg?height=593&width=673&top_left_y=1965&top_left_x=700)

The longer leg in each of these three triangles coincides with side $B C$, which has a length of $6 \mathrm{~cm}$. Additionally, the ratio of the hypotenuse to the shorter leg is $2: 1$. If we denote the length of the hypotenuse by $a$, then according to the Pythagorean theorem, we have

$$
a^{2}=36+\left(\frac{a}{2}\right)^{2}
$$

After simplification, we get $a^{2}=48$, so $a=4 \sqrt{3} \doteq 6.93(\mathrm{~cm})$. This is the length of segment $E F$.

Evaluation. 2 points for determining the necessary angles, or recognizing the halves of equilateral triangles; 2 points for calculating the length of $E F$; 2 points for clarity and quality of the commentary.

Notes. The final calculation can be replaced by referring to the known relationship between the height and the side of an equilateral triangle. With the given notation, we have $6=\frac{\sqrt{3}}{2} a$, so $a=4 \sqrt{3}$.

With knowledge of Thales' theorem, or the converse theorem, the size of angle $A C B$ can be derived as follows: Since the angles at vertices $B$ and $D$ are right angles, both lie on a circle with diameter $A C$. The center of this circle is the midpoint of segment $A C$ and is denoted by $G$ (the same point as mentioned in the solution above). According to the problem, the radius of this circle is equal to the length of side $B C\left(\frac{1}{2}|A C|=|B C|\right)$. Therefore, triangle $G C B$ is equilateral, and in particular, the internal angle at vertex $C$ has a measure of $60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_5c762d128fb89da6abfdg-4.jpg?height=594&width=668&top_left_y=1265&top_left_x=700)

From the above, it follows that points $A, B, C, D$ are four of the vertices of a regular hexagon.