2019 AMC 10B Problem 16

Scale the figure so that $AC=CD=4$ and $DE=EB=3$. Let $A=\angle BAC$ and $B=\angle ABC$, so $A+B=90^\circ$. The isosceles triangles $ACD$ and $DEB$ give $\angle CDA=A$ and $\angle EDB=B$. Because $DA$ and $DB$ are opposite rays, $$\angle CDE=180^\circ-A-B=90^\circ.$$ Therefore $CE=\sqrt{4^2+3^2}=5$, so $BC=CE+EB=8$ and $\tan A=\frac{BC}{AC}=2$. The bases of the two isosceles triangles have lengths $AD=8\cos A$ and $BD=6\cos B=6\sin A$. Hence $$\frac{AD}{BD}=\frac{8\cos A}{6\sin A}=\frac{4}{3\tan A}=\frac23.$$ Thus, the answer is A .