Let $ABCD$ be an isosceles trapezoid with $\overline{BC} \parallel \overline{AD}$ and $AB=CD$. Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y$. Suppose $\angle AXD = \angle BYC = 90^\circ$, $AX = 4$, $XY = 2$, and $YC = 3$. What is the area of $ABCD$?

1. Place $A=(0,0), C=(9,0), X=(4,0), Y=(6,0)$ (so $AX=4,XY=2,YC=3$). Let $D=(4,-p)$, $B=(6,q)$ with $p,q>0$.
2. $AD\parallel BC$ forces $\dfrac pq=\dfrac{AX}{YC}=\dfrac43$, so $p=\dfrac{4q}3$.
3. $AB=CD$ gives $(AX+XY)^2+q^2=(YC+XY)^2+p^2$, i.e.\ $36+q^2=25+p^2$; substituting: $36+q^2=25+\frac{16}9q^2$, so $q^2=\frac{99}7$, $p^2=\frac{176}7$, and $p+q=\sqrt{77}$.
4. $[ABCD]=\frac12\cdot AC\cdot(p+q)=\frac12\cdot9\cdot\sqrt{77}=\boxed{\dfrac{9\sqrt{77}}2}$.