Question

which of the following proportions can be used to prove ae || dk in the proof below? select all that apply.
□ \\( \frac { d a } { d f } = \frac { e f } { k f } \\)
□ \\( \frac { d f } { d a } = \frac { k f } { k e } \\)
□ \\( \frac { f k } { f e } = \frac { f a } { f d } \\)
□ \\( \frac { d a } { a f } = \frac { k e } { e f } \\)
□ \\( \frac { a f } { d a } = \frac { e f } { k e } \\)

Explanation:

Step1: Recall the basic proportionality theorem (Thales' theorem)

If a line divides two sides of a triangle proportionally, then it is parallel to the third side.

Step2: Analyze each proportion

• For \(\frac{DA}{AF}=\frac{KE}{EF}\), by cross - multiplying we get \(DA\times EF = AF\times KE\). Then, \(\frac{DA}{DK}=\frac{AE}{DK}\) (by re - arranging and using segment addition \(DK=DA + AF\) and \(FK = KE+EF\)). This proportion can be used to prove \(AE\parallel DK\) as it satisfies the converse of the basic proportionality theorem.
• For \(\frac{AF}{DA}=\frac{EF}{KE}\), cross - multiplying gives \(AF\times KE=DA\times EF\). By re - arranging and using segment addition \(DK = DA + AF\) and \(FK=KE + EF\), this proportion also satisfies the converse of the basic proportionality theorem.

Answer:

\(\frac{DA}{AF}=\frac{KE}{EF}\), \(\frac{AF}{DA}=\frac{EF}{KE}\)