Question

In the above quadrilateral ABCD,

AD || EF || BC and
|AE| = 2 EB, BM = 4, |AD = 6, BC| =
, 9
Then DO​

Answer 1

The length of segment
`\overline{DO}` is
`\displaystyle (24)/(5)`.

Explanation:

Start by verifying that
`\triangle DAB \sim \triangle MEB`, and that
`\triangle DOA \sim \triangle BOC`.

Both
`\triangle DAB` and
`\triangle MEB` contain the angle
`\angle DBA`. Hence, another pair of equal angles will be sufficient to prove
`\triangle DAB \sim \triangle MEB` by the angle-angle similarity postulate.

Notice that because
`\overline{AD} \; \| \; \overline{EF}`, the corresponding angles next to those two lines would be equal:
`\angle MEB = \angle DAB`. Both
`\angle MEB` is in
`\rm \triangle MEB` whereas
`\angle DAB` is in
`\triangle DAB`.

Hence,
`\triangle DAB \sim \triangle MEB` by the angle-angle similarity postulate.

Notice how
`\angle DOA` in
`\triangle DOA` and
`\angle BOC` in
`\triangle BOC` are two vertically opposite angles (and are thus equal to one another.) That is:
`\angle DOA = \angle BOC`.

Because
`\overline{AD} \| \overline{BC}`, the alternate interior angles
`\angle DAO` and
`\angle BCO` are also equal to one another. That is:
`\angle DAO = \angle BCO`

Hence,
`\triangle DOA \sim \triangle BOC`, also by the angle-angle similarity postulate.

Notice that
`\triangle DAB \sim \triangle MEB` implies that:
`\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 $\left|\overline{AB}\right| = 2\,\left|\overline{EB}\right|$})\\ &= (1)/(3)\end{aligned}`.

Therefore:

`\begin{aligned}&\left|\overline{BD}\right|\\ &= 3\, \left|\overline{BM}\right| \\ &= 12\end{aligned}`.

Similarly,
`\triangle DOA \sim \triangle BOC` implies that:

`\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) \\ &= (6)/(9) = (2)/(3)\end{aligned}`.

Therefore,
`\displaystyle \left|\overline{BO}\right|= (3)/(2)\, \left|\overline{DO}\right|`.

`\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|} = (2)/(5)\end{aligned}`.

It was already found that
`\left|\overline{BD}\right| = 12`. Therefore:

`\displaystyle \left|\overline{DO}\right| = (2)/(5)\, \left|\overline{BD}\right| = (24)/(5) = 4.8`.