Question

can the triangles be proven similar using the sss or sas similarity theorems? yes, △efg ~ △klm only by sss. yes, △efg ~ △klm only by sas. yes, △efg ~ △klm by sss or sas. no, they cannot be proven similar by sss or sas.

Explanation:

Step1: Check SAS similarity

• Calculate the ratios of the sides including the given angles.
• For \(\triangle EFG\) and \(\triangle KLM\), the ratio of the sides adjacent to the equal angles: \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5} = 3\). Since \(\frac{9}{4}

eq3\), SAS similarity does not hold.

Step2: Check SSS similarity

• Calculate the ratios of all three pairs of corresponding sides.
• \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5}=3\), \(\frac{EG}{KM}\) (assuming we consider all - side ratios, but actually, if we re - check the correct side - to - side correspondence for the angles. The correct side - to - side ratios for the angles (using the angle - included sides):
• Let's assume the included - angle sides: For the equal angles, in \(\triangle EFG\) and \(\triangle KLM\), if we consider the sides adjacent to the equal angles.
• Wait, correct approach:
• The ratio of the sides: \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5} = 3\), \(\frac{EG}{KM}\) (but actually, if we use the correct side - pairing for the SAS (angle - included sides).
• Wait, no, for the given angles (the non - marked sides, assume the angles are equal). Let's use the formula for SAS similarity (two sides in proportion and included angle equal).
• The sides adjacent to the equal angles:
• \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5}=3\). Not equal.
• For SSS similarity, we need \(\frac{EF}{KL}=\frac{FG}{LM}=\frac{EG}{KM}\). But \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5} = 3\).
• Wait, wrong. Let's re - do:
• If we assume \(\angle F=\angle L\) (the marked angles).
• Then for SAS similarity: \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5}=3\). Not equal.
• For SSS similarity: We need to check all three sides. But we are not given the third side of \(\triangle KLM\). Wait, no, wait the problem is about SSS or SAS.
• Wait, actually, if we use the ratio of sides:
• \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5}=3\), \(\frac{EG}{KM}\) (but if we assume the sides:
• Wait, no, correct:
• For the two triangles \(\triangle EFG\) and \(\triangle KLM\) with \(\angle F=\angle L\) (marked).
• The ratio of the sides adjacent to the equal angles:
• \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5}=3\). Not equal for SAS.
• For SSS, we need \(\frac{EF}{KL}=\frac{FG}{LM}=\frac{EG}{KM}\). But \(\frac{EF}{KL}=\frac{18}{8}=\frac{9}{4}\), \(\frac{FG}{LM}=\frac{15}{5}=3\). Not equal.

Answer:

No, they cannot be proven similar by SSS or SAS.