Question

can the triangles be proven similar using the sss or sas similarity theorems? yes, \\( \triangle efg \sim \triangle klm \\) only by sss. yes, \\( \triangle efg \sim \triangle klm \\) only by sas. yes, \\( \triangle efg \sim \triangle klm \\) by sss or sas. no, they cannot be proven similar by sss or sas.

Explanation:

Step1: Check the ratio of sides for SAS similarity

For SAS similarity, we need to check if the ratio of two sides of one triangle is equal to the ratio of two sides of the other triangle and the included angles are equal.
The ratio of \( \frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \( \frac{FG}{LM}=\frac{15}{5} = 3\). The ratios are not equal.
Now check another pair: \( \frac{EF}{LM}=\frac{18}{5}\), \( \frac{FG}{KL}=\frac{15}{8}\).
Check \( \frac{EG}{KM}\) and \( \frac{EF}{KL}\): \( \frac{EG}{KM}\) is not given for one triangle. But check the included - angle case.
The included angles (the non - marked angles for side - angle - side) are not given to be equal. Now check SSS.

Step2: Check the ratio of sides for SSS similarity

\(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5} = 3\), \(\frac{EG}{KM}\) (assuming \(KM\) is related). Wait, no.
Wait, re - calculate the ratios correctly.
\(\frac{EF}{KM}=\frac{18}{6}=3\), \(\frac{FG}{LM}=\frac{15}{5} = 3\), \(\frac{EG}{KL}=\frac{24}{8}=3\)
Since \(\frac{EF}{KM}=\frac{FG}{LM}=\frac{EG}{KL}=3\) (by SSS similarity theorem: If the ratios of the corresponding sides of two triangles are equal, then the triangles are similar)

Answer:

Yes, \(\triangle EFG\sim\triangle KLM\) only by SSS.