4. Let $A B C D$ be a cyclic quadrilateral with $A B=3, B C=2, C D=2, D A=4$. Let lines perpendicular to $\overline{B C}$ from $B$ and $C$ meet $\overline{A D}$ at $B^{\prime}$ and $C^{\prime}$, respectively. Let lines perpendicular to $\overline{A D}$ from $A$ and $D$ meet $\overline{B C}$ at $A^{\prime}$ and $D^{\prime}$, respectively. Compute the ratio $\frac{\left.\mid B C C^{\prime} B^{\prime}\right]}{\left[D A A^{\prime} D^{\prime}\right]}$, where $[\varpi]$ denotes the area of figure $\varpi$.
$X=B C \cap A D$, which lies on ray $\overrightarrow{B C}$ and $\overrightarrow{A D}$ since $\widehat{A B}>\widehat{C D}$ (as chords $A B>C D$ ).
By similar triangles we have equality of ratios $X C: X D: 2=(X D+4):(X C+2): 3$, so we have a system of linear equations: $3 X C=2 X D+8$ and $3 X D=2 X C+4$, so $9 X C=6 X D+24=4 X C+32$ gives $X C=\frac{32}{5}$ and $X D=\frac{84 / 5}{3}=\frac{28}{5}$.
It's easy to compute the trapezoid area ratio $\frac{\left[B C^{\prime} B^{\prime} C\right]}{\left[A A^{\prime} D^{\prime} D\right]}=\frac{B C\left(C B^{\prime}+B C^{\prime}\right)}{A D\left(A D^{\prime}+D A^{\prime}\right)}=\frac{B C}{A D} \frac{X C+X B}{X A+X D}$ (where we have similar right triangles due to the common angle at $X$ ). This is just $\frac{B C}{A D} \frac{X C+B C / 2}{X D+A D / 2}=\frac{2}{4} \frac{32 / 5+1}{28 / 5+2}=\frac{37}{76}$.