Answer:
The length of segment [tex]\overline{DO}[/tex] is [tex]\displaystyle \frac{24}{5}[/tex].
Step-by-step explanation:
Start by verifying that [tex]\triangle DAB \sim \triangle MEB[/tex], and that [tex]\triangle DOA \sim \triangle BOC[/tex].
Both [tex]\triangle DAB[/tex] and [tex]\triangle MEB[/tex] contain the angle [tex]\angle DBA[/tex]. Hence, another pair of equal angles will be sufficient to prove [tex]\triangle DAB \sim \triangle MEB[/tex] by the angle-angle similarity postulate.
Notice that because [tex]\overline{AD} \; \| \; \overline{EF}[/tex], the corresponding angles next to those two lines would be equal: [tex]\angle MEB = \angle DAB[/tex]. Both [tex]\angle MEB[/tex] is in [tex]\rm \triangle MEB[/tex] whereas [tex]\angle DAB[/tex] is in [tex]\triangle DAB[/tex].
Hence, [tex]\triangle DAB \sim \triangle MEB[/tex] by the angle-angle similarity postulate.
Notice how [tex]\angle DOA[/tex] in [tex]\triangle DOA[/tex] and [tex]\angle BOC[/tex] in [tex]\triangle BOC[/tex] are two vertically opposite angles (and are thus equal to one another.) That is: [tex]\angle DOA = \angle BOC[/tex].
Because [tex]\overline{AD} \| \overline{BC}[/tex], the alternate interior angles [tex]\angle DAO[/tex] and [tex]\angle BCO[/tex] are also equal to one another. That is: [tex]\angle DAO = \angle BCO[/tex]
Hence, [tex]\triangle DOA \sim \triangle BOC[/tex], also by the angle-angle similarity postulate.
Notice that [tex]\triangle DAB \sim \triangle MEB[/tex] implies that:[tex]\begin{aligned} & \frac{\left| \overline{BM} \right|}{\left| \overline{BD} \right|} \\ &= \frac{\left|\overline{EB}\right|}{\left|\overline{AB}\right|} && (\triangle MEB \sim \triangle DAB) \\ &= \frac{\left|\overline{EB}\right|}{\left|\overline{AE}\right| + \left|\overline{EB}\right|} \\ &= \frac{\left|\overline{EB}\right|}{2\, \left|\overline{EB}\right| + \left|\overline{EB}\right|} && (\text{given that $\left|\overline{AB}\right| = 2\,\left|\overline{EB}\right|$})\\ &= \frac{1}{3}\end{aligned}[/tex].
Therefore:
[tex]\begin{aligned}&\left|\overline{BD}\right|\\ &= 3\, \left|\overline{BM}\right| \\ &= 12\end{aligned}[/tex].
Similarly, [tex]\triangle DOA \sim \triangle BOC[/tex] implies that:
[tex]\begin{aligned} \frac{\left| \overline{DO} \right|}{\left| \overline{BO} \right|} &= \frac{\left|\overline{AD}\right|}{\left|\overline{BC}\right|} && (\triangle DOA \sim \triangle BOC) \\ &= \frac{6}{9} = \frac{2}{3}\end{aligned}[/tex].
Therefore, [tex]\displaystyle \left|\overline{BO}\right|= \frac{3}{2}\, \left|\overline{DO}\right|[/tex].
[tex]\begin{aligned} \frac{\left| \overline{DO} \right|}{\left| \overline{BD} \right|} &= \frac{\left| \overline{DO} \right|}{\left| \overline{BO} \right| + \left| \overline{DO} \right|} \\ &= \frac{\left| \overline{DO} \right|}{(3/2)\, \left| \overline{DO} \right| + \left| \overline{DO} \right|} = \frac{2}{5}\end{aligned}[/tex].
It was already found that [tex]\left|\overline{BD}\right| = 12[/tex]. Therefore:
[tex]\displaystyle \left|\overline{DO}\right| = \frac{2}{5}\, \left|\overline{BD}\right| = \frac{24}{5} = 4.8[/tex].
