geometry - Length of Line in a Square

Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is one and area $XYZ = \frac{2}{3}$ what is the length of $YZ$.

I worked the area of trapezium $ABYZ$ to equal $YZ$:

$\text{Area of square not covered by triangle} = (1-YZ)(1)$

reason ($\text{area of rectangle}=\text{Length}\times \text{Breadth}$)

Therefore $\text{area of trapezium} = 1-(\text{Area of rectangle}) = 1-(1-YZ) = YZ$.

$\text{Area of trapezium} =\left(\frac{a+b}{2}\right)h$

Therefore: $YZ = \frac{(1+YZ)}{2} (h=1)$

$YZ = 1$.

Where did I go wrong.

EDIT please do not give the answer

Answer

Here is your mistake: You worked out the area of the trapezoid by the formula for the area of the rectangle. However, a trapezoid is not a rectangle, so the area of the trapezoid is not equal to YZ.

You were in the right direction though. One way to go about it would be to subtract from 1 the area of the triangles ADY and BCZ... Try it :)