Geometry question

In square ABCD, AD is 4 centimeters, and M is the midpoint of CD.
Let O be the intersection of AC and BM.
Let P be the intersection of DO and BC.
What is the ratio of BP to PC?

My attempt:

$\text{Let $BP = x,~ PC = 4-x,~D=(0,0)$}$

$\begin{array}{|rcll|} \hline \text{Line AC:} \quad y = 4-x \\ \text{Line BM:} \quad y = 2x-4 \\ \hline 2x-4 &=& 4-x \\ 3x &=& 8 \\ \mathbf{ x_{\text{O}} } &=& \mathbf{ \dfrac{8}{3} } \\ \hline y_{\text{O}} &=& 4-\dfrac{8}{3} \\ \mathbf{ y_{\text{O}} } &=& \mathbf{ \dfrac{4}{3} } \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \dfrac{x_{\text{O}}}{y_{\text{O}}} &=& \dfrac{4}{4-x} \\\\ \dfrac{\dfrac{8}{3}}{\dfrac{4}{3}} &=& \dfrac{4}{4-x} \\\\ 2 &=& \dfrac{4}{4-x} \\\\ 2(4-x) &=& 4 \\ 4-x &=& 2 \\ x &=& 4-2 \\ \mathbf{x} &=& \mathbf{2} \\ \mathbf{4-x} &=& \mathbf{4-2=2} \\ \hline \end{array}$

$\dfrac{BP}{PC} = \dfrac{x}{4-x} = \dfrac{2}{2}$