Three parallel lines and segments

Question

On the given diagram $AD\parallel BE\parallel CF, AE=BE=8$ and $BF=CF=16$. What is the length of $AD$? As you can see on the diagram, they have actually found the intersect of $AC$ and $DF$, let’s say $X$. When taking into consideration that point, we have 3 similar triangles, namely $\triangle XAD,\triangle XBE, \triangle XCF$ in which $AD,BE,CF$ are respective sides. I don’t find a way to use this similarity, though. In the context of the given segments maybe? We have $8:16=XE:XF$, but this doesn’t seem helpful.

Answer 1

Here's your diagram with the specified line lengths shown, and several angles specified,

With $AD \parallel BE$, due to the corresponding angles being congruent, we have

$$\measuredangle ADE = \measuredangle BEF = \beta \tag{1}\label{eq1A}$$

Also, with the $3$ parallel lines, and assigning the point where the $2$ outer lines meet as being $X$ as you did (and is shown in the diagram), we have

$$\measuredangle DAX = \measuredangle EBX = \measuredangle FCX = \alpha \tag{2}\label{eq2A}$$

Since $\lvert AE\rvert = \lvert BE\rvert = 8$, then $\triangle AEB$ is isoceles, so $\measuredangle EAB = \measuredangle EBX = \alpha$. Also, $\lvert BF\rvert = \lvert CF\rvert = 16$ means $\triangle BFC$ is isosceles, with $\measuredangle FBC = \measuredangle FCX = \alpha$. Thus,

$$\measuredangle DAE = \measuredangle EBF = \pi - 2\alpha \tag{3}\label{eq3A}$$

Therefore, \eqref{eq1A} and \eqref{eq3A} show by AAA that $\triangle DAE \sim \triangle EBF$, which means

$$\frac{\lvert AD\rvert}{\lvert AE\rvert} = \frac{\lvert BE\rvert}{\lvert BF\rvert} \;\;\to\;\; \frac{\lvert AD\rvert}{8} = \frac{8}{16} \;\;\to\;\; \lvert AD\rvert = 4 \tag{4}\label{eq4A}$$