A6. On the side $A B$ of a square $A B C D$ with $|A B|=3$ lies a point $E$ such that $|A E|=1$ and $|E B|=2$. $A C$ and $D E$ intersect at the point $H$. What is the area of triangle $C D H$?
A) $\frac{9}{8}$
B) 2
C) $\frac{21}{8}$
D) 3
E) $\frac{27}{8}$

![](https://cdn.mathpix.com/cropped/2024_04_17_17b311164fb12c8a5afcg-2.jpg?height=301&width=314&top_left_y=516&top_left_x=1594)
A6. E) $\frac{27}{8}$ Draw a line through $H$ parallel to $A D$. Name the intersection points of this line with $A B$ and $C D$ as $P$ and $Q$ respectively. Now, $|H P|:|H Q|=|A E|:|C D|=1: 3$, so $|H Q|=\frac{3}{4} \times|P Q|=\frac{3}{4} \times 3=\frac{9}{4}$. The area of triangle $C D H$ is thus $\frac{1}{2} \times 3 \times \frac{9}{4}=\frac{27}{8}$.

![](https://cdn.mathpix.com/cropped/2024_04_17_7fefb8cd27a15e1d29a5g-1.jpg?height=317&width=328&top_left_y=2335&top_left_x=1560)