Prove that EF || BC and 4X = AF imes AB?

Here's a step-by-step solution to prove $EF || BC$ and address the likely intended first part of the question.

The problem statement contains a likely typo in the first part "Prove that $4X = AF \cdot AB$". This expression is dimensionally inconsistent (length equals area). Given the context of proving $EF || BC$, which involves ratios of lengths, it's highly probable that the first part also relates to ratios or lengths involving $AX$, $AD$, $AO$, $OD$, $AF$, $AB$. A common related property in such configurations is $AX = \frac{k+2}{k+1} AD$ where $AO = k \cdot OD$. If $O$ is the centroid, $k=2$, leading to $AX