Since we're working on a rectangle, we know that the diagonals intercect right in the middle, and that they're all equal.
This way,
![AC=BD=22](https://img.qammunity.org/2023/formulas/mathematics/college/ndqqbajuqen6yssyhir17qmxq0nzoy5glf.png)
And
![AE=BE=DE=CE=(1)/(2)AC=11](https://img.qammunity.org/2023/formulas/mathematics/college/jc8s0mctjavpapksmvul3d9uxyzqm72r2m.png)
Now, we also know that
![AD=BC=9](https://img.qammunity.org/2023/formulas/mathematics/college/lupegpd18of496h6hqzkpjjf1f03brx6gc.png)
And using the Pythagorean theorem, we have that:
![AD^2+CD^2=AC^2](https://img.qammunity.org/2023/formulas/mathematics/college/rvziuli0jqveu8zsyduxn8y3rld6yh990l.png)
Solving for CD,
![\begin{gathered} CD=\sqrt[]{AC^2-AD^2} \\ \rightarrow CD=20.07 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zf4t8i9jjk9epccgk91m186mttubnamwpa.png)
And we know that:
![AB=DC=20.07](https://img.qammunity.org/2023/formulas/mathematics/college/v4nxeruw59eyafvgsn7sncvtffx06zfpdd.png)
This way, the final answers would be:
![\begin{gathered} AD=9 \\ AC=22 \\ CD=20.07 \\ AB=20.07 \\ BD=22 \\ BC=9 \\ AE=11 \\ BE=11 \\ CE=11 \\ DE=11 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zg7z7s5977okx0qrbat9gfhmmyuvowc9fj.png)