Prove $AC=\frac12AD$ in the triangle with given conditions. The question reads:

Prove the following: if in triangle $ABC$, median $AM$ is such that $\text m\angle BAC$ is divided into the ratio $1:2$ and $AM$ is extended through $M$ to $,D$ so that $\angle DBA$ is a right angle, then $AC = \frac12AD$.

So far I have determined that the two angles must be $30^\circ$ and $60^\circ$. Also I tried making a rectangle and seeing if I could prove something that way, but each method reaches a dead end. Could anyone please provide a HINT, so that I am able to solve this problem with some progress? All help is appreciated!
 A: Let $\angle{BAM}=\alpha, \angle{MAC}=2\alpha$.
Let $\overline{AC}=x, \overline{AD}=y, \overline{AM}=m$. Then, $\overline{AB}=y\cos{\alpha}$.
$\text{Area}(\triangle{BAM})=\frac{1}{2}(y\cos{\alpha})(m)\sin{\alpha}$
$\text{Area}(\triangle{MAC})=\frac{1}{2}(m)(x)\sin{2\alpha}$
Since $AM$ is a median, these two areas are equal. Equate them, and the result follows.

An entirely geometric solution is as follows: It makes use of the (given) fact that $\angle{BAC}=90^\circ$.
Construct the quadrilateral $ABDC$. This is a rectangle, since $AD$ bisects $BC$ (and we already have $\angle{BAC}=\angle{ABD}=90^\circ$).
Next, $\angle{MAC}=60^\circ$, so $\triangle{AMC}$ is equilateral (use the rectangle part). Thus, $AC=AM=MC$. Since the diagonals of a rectangle bisect at intersection, $AM=MD$. So, $AD=2AM=2AC$.
A: Construct the parallelogram $ABNC$, with $\widehat{CAN}=2\widehat{BAN}$, the intersection of $AN$ with $BC$ gives $M$. Take $P$ midpoint of $AD$, you get $\widehat{BPD}=2\widehat{BAN}$, implying $BP=BN=AC$, done.
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