Let $A B C D$ and $E F G H$ be squares with side lengths 33 and 12, respectively, with $E F$ lying on side $D C$ (as shown in the figure below). Let $X$ be the intersection point of segments $H B$ and $D C$. Suppose that $\overline{D E}=18$.

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-17.jpg?height=754&width=542&top_left_y=1209&top_left_x=839)

a) Calculate the length of the segment $\overline{E X}$.

b) Prove that the points $A, X$, and $G$ are collinear.
Solution

a) Denote $\overline{E X}=x$. We have that $|\overline{C X}|=33-18-x=15-x$.

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-54.jpg?height=999&width=727&top_left_y=471&top_left_x=752)

Now note that the triangles $E X H$ and $C X B$ are similar, so:

$$
\frac{|\overline{E H}|}{|\overline{C B}|}=\frac{|\overline{E X}|}{|\overline{C X}|} \Rightarrow \frac{12}{33}=\frac{x}{15-x}
$$

Now we find $x$ :

$$
\begin{aligned}
12(15-x) & =33 x \\
4(15-x) & =11 x \\
60-4 x & =11 x \\
15 x & =60 \\
x & =4
\end{aligned}
$$

Therefore, $|\overline{E X}|=4$.

b) Let $Y$ be the intersection of the line $A G$ with the segment $D C$. To prove that $A, X$ and $G$ are collinear, it suffices to show that $Y=X$. And for that, we will calculate $|\overline{E Y}|$, and then see that this is equal to $|\overline{E X}|$. Denote $|\overline{E Y}|=y$. Therefore, $|\overline{F Y}|=12-y$.

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-55.jpg?height=994&width=724&top_left_y=193&top_left_x=550)

Analogously to the previous case, we see that the triangles $F Y G$ and $D Y A$ are similar. Therefore:

$$
\frac{|\overline{F G}|}{|\overline{D A}|}=\frac{|\overline{F Y}|}{|\overline{D Y}|} \quad \Rightarrow \frac{12}{33}=\frac{12-y}{18+y}
$$

Now we find $y$ :

$$
\begin{array}{ccc}
12(18+y) & = & 33(12-y) \\
4(18+y) & = & 11(12-y) \\
72+4 y & = & 132-11 y \\
15 y & = & 60 \\
y & = & 4 .
\end{array}
$$

Therefore, $|\overline{E Y}|=|\overline{E X}|$. Thus, $X=Y$, and we conclude that the points $A, X$ and $G$ are collinear.