Consider the trapezoid $ABCD$ with bases $BC$ and $AD$ such that $AB = BC = CD = 5$ and $AD = 10$. Let $E$ be the intersection point of the diagonals $AC$ and $BD$. The line perpendicular to $AC$ drawn through $E$ intersects the extension of $AB$ at $F$ and the base $AD$ at $H$.

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-10.jpg?height=553&width=870&top_left_y=289&top_left_x=485)

a) Determine the length of $AH$.

b) Determine the length of $AE$.

c) Find the area of the quadrilateral $AFCH$.
Solution

a) Initially, we will verify that $ABCD$ is half of a regular hexagon. Let $M$ be the midpoint of $AD$. Since $BC$ and $AM$ are equal and parallel, $ABCM$ is a parallelogram. Furthermore, since $AM=AB=BC$, it follows that $CM=AB=CD=DM$. Thus, $CDM$ is an equilateral triangle. Similarly, we can obtain $BM=CM=CD$. Therefore, the triangles $ABM$, $BCM$, and $CDM$ are congruent, and the circle with center $M$ and radius $CM$ passes through $A$, $B$, $C$, and $D$. Hence,

$$
\angle BAD=60^{\circ} \text{ and } \angle ABC=180^{\circ}-\angle BAM=120^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_01_82072e76ed187b592b1cg-10.jpg?height=288&width=576&top_left_y=1791&top_left_x=627)

From $AB=BC$ it follows that $\angle BAC=\angle BCA=\angle CAD$. Thus, $\angle BAC=\angle CAD=30^{\circ}$. Since $FH \perp AE$, we have $\triangle AFH$ equilateral. Moreover, $\angle ACD=\angle DBA=90^{\circ}$. Therefore, $EH \| CD$. Since the triangles $\triangle BEC$ and $\triangle AED$ are similar, we have

$$
\frac{AH}{HD}=\frac{AE}{EC}=\frac{10}{5}=2
$$

Consequently, $HD=10 / 3$ and $AH=20 / 3$.
b) In the right triangle $AEH$, we have

$$
AE=AH \cdot \cos 30^{\circ}=\frac{20}{3} \cdot \frac{\sqrt{3}}{2}=\frac{10 \sqrt{3}}{3}
$$

c) Since the diagonals of $AFCH$ are perpendicular, we have $[AFCH]=\frac{AC \cdot FH}{2}$. By the Pythagorean theorem, it follows that

$$
AC=\sqrt{AD^{2}-CD^{2}}=\sqrt{100-25}=5 \sqrt{3}
$$

Another way to obtain the length of this segment is to calculate

$$
AC=AD \cdot \cos 30^{\circ}=5 \sqrt{3}
$$

Thus,

$$
\begin{aligned}
A_{AFCH} & =\frac{AC \cdot FH}{2} \\
& =\frac{AC \cdot AH}{2} \\
& =\frac{5 \sqrt{3} \cdot 20 / 3}{2} \\
& =\frac{50 \sqrt{3}}{3}
\end{aligned}
$$