Note that $\angle CAB$ is obtuse iff $A$ lies inside the circle with diameter $BC$, i.e. $DC<AD$. If $AC<2AD$, then there exists a triangle such that $AD=DC$, so $\angle CAB$ is not obtuse, but right, so $AC\geq 2AD$. Now let $E$ be the midpoint of $AC$. Then $\angle BAD=\angle ADE$. If $AC>2AD$, then $AE>AD$, so there exists a triangle such that $\angle ADE=90$, thus $\angle BAD$ is not acute. Thus we have $AC\leq 2AD$. Combining both inequalities, we have $AC=2AD$, and it is easy to show that any triangle with $\frac{AD}{AC}=\frac12$, satisfies the conditions using similar method.