imc

Without loss of generality, let $AC=CD=4$ and $DE=EB=3$. Let $\angle A=\alpha$ and $\angle B=\beta =90^{\circ }-\alpha$. As $△ACD$ and $△DEB$ are isosceles, $\angle ADC=\alpha$ and $\angle BDE=\beta$. Then $\angle CDE=180^{\circ }-\alpha -\beta =90^{\circ }$, so $△CDE$ is a $3-4-5$ triangle with $CE=5$. Then $CB=5+3=8$, and $△ABC$ is a $1-2-\sqrt{5}$ triangle. In isosceles triangles $△ACD$ and $△DEB$, drop altitudes from $C$ and $E$ onto $AB$; denote the feet of these altitudes by $P_C$ and $P_E$ respectively. Then $△AC{P}_C\sim △ABC$ by AAA similarity, so we get that $A{P}_C=P_{C}D=\frac{4}{\sqrt{5}}$, and $AD=2\times \frac{4}{\sqrt{5}}$. Similarly, we get $BD=2\times \frac{6}{\sqrt{5}}$, and $AD:DB=\boxed{2:3}$.