Let $A'$ be a point such that $ABA'C$ is parallelogram ,thus, $AB=AD=DC=CA'$ that means $\angle CDA'=\angle CA'D=90-\angle ADB=90-\angle ABD$. By taking intersection point of $DA'$ with $BC$ (call it $E$) and midpoint of $BC$ (call it $M$) we got harmonic points $(B,E;M,C)=-1$. So, $\frac{BM}{BA}=\frac{EM}{MC}=\frac{1}{2}$. Which gives us desired condition.