23) In the square $A B C D$, with a side length of $12 \mathrm{~m}$, the side $A B$ is divided into three segments $A E, E F$ and $F B$ of equal length. The segments $E C$ and $F D$ are drawn, intersecting at point $H$. What is the area of triangle $H C D$?
(A) $36 \mathrm{~m}^{2}$,
(B) $48 \mathrm{~m}^{2}$
(C) $54 \mathrm{~m}^{2}$,
(D) $60 \mathrm{~m}^{2}$
(E) $72 \mathrm{~m}^{2}$.

## 00020001 is:
23. The answer is $\mathbf{( C )}$.

Triangle $C D H$ is isosceles. If $H K$ is its altitude relative to $C D$, then $\overline{D K}=\overline{K C}$. Let $G$ be the intersection point between the extension of $B C$ and the extension of $D F$. Triangles $D H K$ and $D G C$ are similar, so $\overline{H K}=\frac{1}{2} \overline{C G}$. Triangles $F B G$ and $D C G$ are also similar, so the proportion $\overline{D C}: \overline{F B} = \overline{C G}: \overline{B G}$ holds, from which $3 \overline{B G}=\overline{C G}$. Furthermore, $\overline{C G}=\overline{B C}+\overline{B G}$, and thus $\overline{C G}=\frac{3}{2} \overline{B C}=18 \mathrm{~m}$ and $\overline{H K}=\frac{1}{2} \overline{C G}=9 \mathrm{~m}$. Finally,

![](https://cdn.mathpix.com/cropped/2024_04_17_4701f996ee50400fb9c5g-5.jpg?height=600&width=428&top_left_y=2030&top_left_x=1499)