The figure shows congruent right triangles ADH and FGE and squares
AHJK, ABCD and DEGH. What is the ratio of the area of
quadrilateral CDHJ to the area of hexagon ADEFGH?
Express your answer as a common fraction.
First, let's label the side lengths of our shapes. Let $AD = DH = a$, $FG = GE = a$, $AH = FG = GE = a$, and $AB = BC = CD = DA = a$.
Now, let's look at the area of quadrilateral $CDHJ$. We can split it up into two triangles: $\triangle CDH$ and $\triangle CJH$. The area of a triangle is $\frac{1}{2} \text{{ base }} \times \text{{ height }}$.
The base of both triangles is $CD = a$, and the height of both triangles is $CH$. Let's look at triangle $\triangle CHJ$. We know that triangle $CHJ$ is congruent to triangle $ADE$. We know that the area of triangle $ADE$ is $\frac{1}{2} \times AD \times DE$, or $\frac{1}{2} \times a \times a = \frac{a^2}{2}$. Since triangle $\triangle CHJ$ is congruent to triangle $ADE$, it also has the area of $\frac{a^2}{2}$. Therefore, the total area of quadrilateral $CDHJ$ is $2 \times \frac{a^2}{2} = a^2$.
Now, let's look at the area of hexagon $ADEFGH$. We can split it up into three triangles: $\triangle ADH$, $\triangle DEH$, and $\triangle FEG$.
The area of triangle $\triangle ADH$ is $\frac{1}{2} \times AD \times DH = \frac{1}{2} \times a \times a = \frac{a^2}{2}$.
The area of triangle $\triangle DEH$ is $\frac{1}{2} \times DE \times EH$. We already know that triangle $\triangle DEH$ is congruent to triangle $\triangle CJH$, so it also has the area of $\frac{a^2}{2}$.
The area of triangle $\triangle FEG$ is $\frac{1}{2} \times FE \times EG$. We know that triangle $FEG$ is congruent to triangle $ADH$, so it also has the area of $\frac{a^2}{2}$.
Therefore, the total area of hexagon $ADEFGH$ is $\frac{a^2}{2} + \frac{a^2}{2} + \frac{a^2}{2} = \frac{3a^2}{2}$.
Finally, we can find the ratio of the area of quadrilateral $CDHJ$ to the area of hexagon $ADEFGH$ by dividing the area of quadrilateral $CDHJ$ by the area of hexagon $ADEFGH$: $\frac{a^2}{\frac{3a^2}{2}} = \boxed{\frac{2}{3}}$.