Given points $A$, $B$, $C$, $D$ in a straight line, find $O$ in the line such that $OA:OB=OC:OD$. 
Given four points $A$, $B$, $C$, $D$ in a straight line, find a point $O$ in the same straight line such that $OA:OB=OC:OD$.

I tried doing this by drawing a ray through $A$ and drawing lines through $B$, $C$, and $D$ parallel to each other. That way we get three triangles. I do not know what to do from here.

Am I going in the right direction? And if no, provide a hint.

Thanks
 A: 
Make another line parallel to the given line and mark $A',B',C',D'$ vertically above $A,B,C,D$. Connect $A'C$ and $B'D$ and mark the intersection $K$. Make a line through $K$ parallel to $DD'$. 
Since 
$$\Delta A'B'K \simeq \Delta CDK$$
We have
$$O'A':O'B' = OC:OD$$
Notice that
$$O'A'= OA \text{ and } O'B' =OB$$
And we are done.
A: I will play with the case where Larry's great solution fails to find a point (or you can say that Larry's solution yields the point $O$ at infinity, but there are two finite points that also satisfy the condition).  That is, when the midpoint of the line segment $AD$ coincides with the midpoint of the line segment $BC$.  Let $M$ denote this common midpoint.
WLOG, suppose that the points $A,B,M,C,D$ are arranged from left to right on the given line $\ell$.  Let $P$ be the midpoint of $AB$.  Construct a right triangle $XYZ$ with $\angle XYZ=\pi/2$, $XY=AP$, and $YZ=MP$.  Let $O_1,O_2$ be two distinct points on the line $\ell$ that are at distance $XZ$ away from $M$.  Then $O=O_1$ or $O=O_2$ will give you a desired point $O$.
To show these indeed are solutions, we suppose that $AM=a$ and $BM=b$.  Then $XY=MP=(a+b)/2$ and $YZ=AP=(a-b)/2$.  Hence, $$MO_1=MO_2=XZ=\sqrt{XY^2+YZ^2}=\sqrt{\frac{a^2+b^2}2}.$$
WLOG, suppose that $O_1$ is on the left of $M$ and $O_2$ is on the right.  Then
$$\frac{O_1A}{O_1B}=\frac{a-\sqrt{(a^2+b^2)/2}}{\sqrt{(a^2+b^2)/2}-b}=\frac{\frac{(a^2-b^2)/2}{a+\sqrt{(a^2+b^2)/2}}}{\frac{(a^2-b^2)/2}{\sqrt{(a^2+b^2)/2}}+b}=\frac{\sqrt{(a^2+b^2)/2}+b}{a+\sqrt{(a^2+b^2)/2}}.$$
However
$$\frac{O_1C}{O_1D}=\frac{\sqrt{(a^2+b^2)/2}+b}{a+\sqrt{(a^2+b^2)/2}}.$$
We have a similar proof for $O_2$.  So both $O_1$ and $O_2$ satisfy the condition.

In general, let $R_1$ be the midpoint of $AD$, $R_2$ the midpoint of $BC$, and $M$ the midpoint of $R_1R_2$.  Suppose now that  $A,B,C,D$ are arranged from left to right on the given line $\ell$.  Now we measure distance with signs that is $UV=-VU$ and $UV$ is positive if $U$ is on the left of $V$.  Let $AM=a$, $BM=b$, $CM=c$, and $DM=d$.
Then apart from point $O_0$ obtained in Larry's solution, there are two other points $O_1,O_2$ if and only if $ad+bc\le 0$.  In such cases, the other two points satisfies
$$O_1M=-\sqrt{-\frac{ad+bc}{2}}$$
and $$O_2M=\sqrt{-\frac{ad+bc}{2}}.$$
You can construct $\delta=\sqrt{-\frac{ad+bc}{2}}$ via
$$\delta=\sqrt{\left(\frac{a+b-c-d}{4}\right)^2+\left(\frac{a-b+c-d}{4}\right)^2-\left(\frac{a+b+c+d}{4}\right)^2-\left(\frac{a-b-c+d}{4}\right)^2}.$$
Notice that $a+b+c+d=0$.  Let $P_1$ be the midpoint of $AB$, $P_2$ the midpoint of $CD$, $Q_1$ be the midpoint of $AC$, and $Q_2$ the midpoint of $BD$.  Then,
$$\delta=\sqrt{MP_1^2+MQ_1^2-MR_1^2}.$$
We then construct $\delta$ as follows.  First draw a right triangle $XYZ$ with $\angle XYZ=\pi/2$, $XY=|MP_1|$, and $YZ=|MQ_1|$.  Then on the circumcircle of the triangle $XYZ$, locate a point $W$ such that $XW=|MR_1|$.  Then, $\delta=ZW$. 
The points $O_1$ and $O_2$ work simply because the roots of
$$t^2-\left(-\frac{ad+bc}{2}\right)=\frac{(t-a)(t-d)+(t-b)(t-c)}{2}=0$$
are precisely $\pm\delta$.  This means
$$\frac{|t-a|}{|t-b|}=\frac{|t-c|}{|t-d|}$$
for $t=\pm\delta$.  
In some degenerate cases such as when three of the points $A,B,C,D$ coincide, say $B=C=D$, one of the points $O_1$ and $O_2$, say $O_1$, is the midpoint of $AD$.  So we do not have division by $0$ in $\frac{O_1A}{O_1B}=\frac{O_1C}{O_1D}$ but Larry's solution yields $O_0=B=D$.  Hence, it is useful to know how to construct $O_1$ and $O_2$ just in case things fail (and in contest math, sometimes you have to be careful to cover all bases).  In a real contest, since the problem doesn't state that $A,B,C,D$ are necessarily distinct points, we should perhaps include the case $A=B$ and $C=D$ where the point $O$ is any point on the straight line not equal to $A=B$ or $C=D$, just to be careful.
There are exactly three degenerate cases where $O$ does not exist (as a finite point).  In the case when $A=B$ is the midpoint of the segment $CD$, it can be shown that the equality $OA\cdot OD=OB\cdot OC$ holds if and only if $O=A=B$, but then we have division by $0$ in $\frac{OA}{OB}=\frac{OC}{OD}$.  Similarly when $B=D$ is the midpoint of $AC$ or when $C=D$ is the midpoint of $AB$, we have a similar problem.
