What's the ratio between these two lengths? plane geometry problem I'm thinking following plane geometry problem.
Question:
There is a parallelogram $ABCD$ such that $\overline{AC}:\overline{BD}=2:1$ and $\overline{AB}\neq\overline{BC}$. Draw a line which is symmetry of $\overline{AD}$ against $\overline{AC}$ and, Draw a line which is symmetry of $\overline{BC}$ against $\overline{BD}$. Let $M$ be the intersection point of those two lines.
Then, we have a following picture.

Observe that $\angle DAO=\angle MAO$,$\angle CBO=\angle MBO$.
Then, what is $\overline{AM}:\overline{BM}$?
Progress so far: I believe answer is $4:1$ because $\triangle AMO$ and $\triangle OMB$ are similar. But I can't find a way to prove the similarity.
Thanks for any help in advance!
 A: This answer should be more elementary.
$\qquad\qquad\qquad\qquad$
Let $N$ be the intersection point between the symmetric line of $AD$ about $BD$ and the symmetric line of $BC$ about $AC$. Also, let $P$ be the intersection point between the symmetric line of $BC$ about $BD$ and the symmetric line of $BC$ about $AC$. And let $Q$ be the intersection point between the symmetric line of $AD$ about $AC$ and the symmetric line of $AD$ about $BD$.
Now, since $O$ is the incenter of both $\triangle{PBC}$ and $\triangle{QAD}$, we see that $PO,QO$ is the bisector of $\angle{BPC},\angle{AQD}$ respectively. Then, we have
$$\angle{AOB}=\angle{OBC}+\angle{OCB},\quad \angle{POA}=\angle{OPC}+\angle{OCP}$$
$$\angle{POD}=\angle{OPB}+\angle{OBP},\quad \angle{DOC}=\angle{OBC}+\angle{OCB}$$
$$\angle{COQ}=\angle{OAQ}+\angle{OQA},\quad \angle{QOB}=\angle{ODQ}+\angle{OQD}$$
Since $O$ is on $AC$, 
$$2(\text{red $+$ light blue $+$ black})=180^\circ\Rightarrow \text{red $+$ light blue $+$ black}=90^\circ$$
$$2(\text{red $+$ green $+$ black})=180^\circ\Rightarrow \text{red $+$ green $+$ black}=90^\circ$$
and so we have
$$\text{light blue $=$ green}\tag1$$
Hence, we can see that $O$ is on the line $PQ$.
From $(1)$, we have that $PM$ is parallel to $NQ$, and that $PN$ is parallel to $MQ$. Since we have $\triangle{POR}$ and $\triangle{QOS}$ are congruent where $R$ is the intersection point between $PC$ and $BD$, and $S$ is the intersection point between $AQ$ and $BD$, we can have $PO=QO$, and so $O$ is on the line $MN$. 
It follows from above that $$\angle{PMO}=\angle{QMO},$$ from which we have that $\triangle{AMO}$ and $\triangle{OMB}$ are similar, and so the answer is $$AM:BM=\color{red}{4:1}.$$
A: This answer uses vectors.
Let $\vec{OA}=\vec a,\vec{OB}=\vec b$. We may suppose that $$|\vec a|=2,\quad |\vec b|=1,\quad \vec a\cdot\vec b=2\cos\theta$$
where $\theta\ (\not=90^\circ)$ is the angle between $\vec a$ and $\vec b$.
Let $P$ be the point on $AC$ such that $PD$ is perpendicular to $AC$. Also, let $N$ be the point such that $P$ is the midpoint of the line segment $ND$.
As there exists a real number $k\not=0$ such that $\vec{OP}=k\vec a$, we have that
$$\vec{PD}=\vec{NP}\iff -\vec b-k\vec a=k\vec a-\vec{ON}\iff \vec{ON}=2k\vec a+\vec b$$
and that
$$\vec{PO}\cdot\vec{PD}=0\iff -k\vec a\cdot(-\vec b-k\vec a)=0\iff k\cdot 2\cos\theta+k^2\cdot 2^2=0\iff k=-\frac{\cos\theta}{2}.$$
Hence, we have $\vec{ON}=-\cos\theta\ \vec a+\vec b$, and there exists a real number $m$ such that
$$\vec{AM}=m\vec{AN}\iff \vec{OM}-\vec a=m(\vec{ON}-\vec a)\iff \vec{OM}=(1-m-m\cos\theta)\vec a+m\vec b\tag1$$
On the other hand, let $Q$ be the point on $BD$ such that $CQ$ is perpendicular to $BD$. Also, let $R$ be the point such that $Q$ is the midpoint of the line segment $RC$.
As there exists a real nuber $q\not=0$ such that $\vec{OQ}=q\vec b$, we have that 
$$\vec{CQ}=\vec{QR}\iff q\vec b-(-\vec a)=\vec{OR}-q\vec b\iff \vec{OR}=2q\vec b+\vec a$$
and that
$$\vec{QO}\cdot\vec{QC}=0\iff -q\vec b\cdot(-\vec a-q\vec b)=0\iff q\cdot 2\cos\theta +q^2\cdot 1^2=0\iff q=-2\cos\theta.$$
Hence, we have $\vec{OR}=-4\cos\theta\ \vec b+\vec a$, and there exists a real number $s$ such that
$$\vec{BM}=s\vec{BR}\iff \vec{OM}-\vec b=s(\vec{OR}-\vec b)\iff\vec{OM}=s\vec a+(1-s-4s\cos\theta)\vec b\tag2$$
From $(1)(2)$, solving the following system
$$1-m-m\cos\theta=s$$
$$m=1-s-4s\cos\theta$$
 gives
$$s=\frac{1}{4\cos\theta +5},\quad m=\frac{4}{4\cos\theta+5}$$
and so
$$\vec{OM}=\frac{1}{4\cos\theta +5}\vec a+\frac{4}{4\cos\theta+5}\vec b$$
Thus,
$$\vec{AM}=\vec{OM}-\vec a=\frac{-4-4\cos\theta}{4\cos\theta +5}\vec a+\frac{4}{4\cos\theta+5}\vec b$$
$$\vec{BM}=\vec{OM}-\vec b=\frac{1}{4\cos\theta +5}\vec a+\frac{-1-4\cos\theta}{4\cos\theta+5}\vec b$$
Hence, since we have
$$AM^2=\frac{16}{4\cos\theta+5},\quad BM^2=\frac{1}{4\cos\theta+5}$$
the answer is
$$AM : BM=\frac{4}{\sqrt{4\cos\theta+5}} : \frac{1}{\sqrt{4\cos\theta+5}}=\color{red}{4:1}.$$
A: The similarity of the two triangles $AMO$ and $OMB$ can be proven by simple angle chasing, see the following figure.

Let $\theta$ be the angle between $AC$ and $BD$; it follows that $\alpha+\beta=\theta$. Reflect $C$ in the line $BD$ to obtain $C'$, and reflect $B$ in the normal to $AC$ to obtain $B'$ (which is the same as reflecting $D$ in $AC$). The triangles $OC'B$ and $OAB'$ are then mirror images of the triangle $OCB$, hence properly congruent. In fact $OAB'$ is obtained from $OC'B$ by a counterclockwise rotation around $O$ by the angle $\theta':=\pi-2\theta$. This then implies that $C'\vee B$ and $A\vee B'$ have the same distance from $O$, hence $OM$ bisects the angle between these two lines. Now check that all angles are correctly identified in the figure.
