Question

if \\(\overline{am} = 21\\), \\(\overline{mb} = 9\\), \\(\overline{an} = 14\\), \\(\overline{nc} = 6\\), then the triangles are similar by:
aa similarity postulate.
sss similarity theorem.
sas similarity theorem.
the triangles are not similar.

Explanation:

Step1: Calculate ratios of sides

First, find \( AB = AM + MB = 21 + 9 = 30 \) and \( AC = AN + NC = 14 + 6 = 20 \). Then, calculate the ratios \(\frac{AM}{AC}=\frac{21}{20}\)? Wait, no, wait. Wait, the triangles are \(\triangle AMN\) and \(\triangle ABC\)? Wait, no, let's check the sides. Wait, \( \frac{AM}{AB} = \frac{21}{30}=\frac{7}{10} \)? No, wait, \( AM = 21 \), \( MB = 9 \), so \( AB = AM + MB = 30 \). \( AN = 14 \), \( NC = 6 \), so \( AC = AN + NC = 20 \). Wait, no, maybe the triangles are \(\triangle AMN\) and \(\triangle ABC\) with angle \( A \) common. So check the ratios of the sides around angle \( A \). So \( \frac{AM}{AC} \)? No, wait, \( AM = 21 \), \( AB = 30 \)? Wait, no, maybe I mixed up. Wait, \( AM = 21 \), \( MB = 9 \), so \( AB = AM + MB = 30 \). \( AN = 14 \), \( NC = 6 \), so \( AC = AN + NC = 20 \). Wait, no, the sides adjacent to angle \( A \) are \( AM \) and \( AB \)? No, wait, the triangles are \(\triangle AMN\) and \(\triangle ABC\), with \( \angle A \) common. So the sides around \( \angle A \) are \( AM \) and \( AB \), \( AN \) and \( AC \)? Wait, no, \( AM = 21 \), \( AB = AM + MB = 30 \)? No, that's not right. Wait, maybe \( \triangle AMN \) and \( \triangle ABC \): \( AM = 21 \), \( AB = AM + MB = 30 \)? No, that can't be. Wait, no, let's recalculate the ratios correctly. \( \frac{AM}{AC} = \frac{21}{20} \)? No, wait, \( AM = 21 \), \( AC = 20 \)? No, that's not. Wait, no, I think I made a mistake. Wait, \( AN = 14 \), \( AC = 20 \), so \( \frac{AN}{AC} = \frac{14}{20} = \frac{7}{10} \). \( AM = 21 \), \( AB = 30 \), so \( \frac{AM}{AB} = \frac{21}{30} = \frac{7}{10} \). And the included angle \( \angle A \) is common. So by SAS similarity, because two sides are in proportion and included angle is equal. Wait, let's check: \( \frac{AM}{AB} = \frac{21}{30} = \frac{7}{10} \), \( \frac{AN}{AC} = \frac{14}{20} = \frac{7}{10} \), and \( \angle A \) is common. So the ratio of \( AM \) to \( AB \) and \( AN \) to \( AC \) are equal, and included angle \( A \) is equal. So by SAS similarity theorem. Wait, let's check the ratios again. \( AM = 21 \), \( AB = AM + MB = 21 + 9 = 30 \). \( AN = 14 \), \( AC = AN + NC = 14 + 6 = 20 \). So \( \frac{AM}{AB} = \frac{21}{30} = \frac{7}{10} \), \( \frac{AN}{AC} = \frac{14}{20} = \frac{7}{10} \). So the two sides around \( \angle A \) are in proportion, and \( \angle A \) is common, so SAS similarity.

Step2: Identify similarity criterion

Since the ratio of the two sides ( \( \frac{AM}{AB} = \frac{AN}{AC} \)) around the common angle \( \angle A \) are equal, the triangles are similar by SAS similarity theorem.

Answer:

SAS similarity theorem (the option: SAS similarity theorem)