If ($ACBD$) $= -1$, show that $DA \cdot DB = DC \cdot DO$, where O is the midpoint of the segment $AB$. 
If ($ACBD$) $= -1$, show that $DA \cdot DB = DC \cdot DO$, where O is the midpoint of the segment $AB$.

My attempt to proof:

Suppose ($ACBD$) $= -1$ then $\frac{AC}{CB} = \frac{AD}{BD}$ then $AC \cdot BD = CB \cdot AD$. Also, since $O$ is a midpoint for $AB$  then $AO = OB$.
\begin{align*}
AC \cdot BD &= CB \cdot AD \\
\Longrightarrow (AD + DC) \cdot BD &= (CD + DB) \cdot(AO + OD) \\
\Longrightarrow AD \cdot BD + BD \cdot DC &= CD \cdot AO + DB \cdot OD + DB \cdot AO + CD \cdot OD \\
\Longrightarrow  AD \cdot BD - CD \cdot OD &= CD \cdot AO + DB \cdot DC + DB \cdot OD + DB \cdot AO \\
&= CD \cdot AO + DB \cdot DC + DB \cdot AD \\
&= CD \cdot AO + DB \cdot AC \\
&= (CA + AD) \cdot AO + (DA + AB) \cdot AC \\
&= CA \cdot AO + AD \cdot AO + DA \cdot AC + AB \cdot AC \\
&= AC \cdot (AB - AO) + DA \cdot (AC - AO) \\
&= AC(OB) + DA(OC)
\end{align*}
I'm stuck in this. Is my proof wrong? how to make the right hand side $0$?

Rephrasing of problem statement

Let $ A C = a, CB = b, BD = c$.
We are given that $ac = (a+b+c)b$.
WTS $(a+b+c)c = (b+c)(c+ (a+b)/2)$.


 Taking the difference of the 2 sides, and multiplying by 2
 WTS $ -ab + ac - b^2 - bc = 0$.
 Notice that this is exactly the condition $ac = (a+b+c)b$.    Hence we are done.


Note: If so desired, you can rephrase this in terms of $AC, CB, BD$, and use only those line segments, to make it clear what we should be focusing on.
Introducing line segments like $OA, OC$ etc makes it harder to manipulate, ad we should just replace them by $OA = (AC+CB)/2$, etc.
