Question

decide whether the triangles are similar. if they are, write a similarity statement and state the reason justifying the similarity
if necessary, you may learn what the markings on a figure indicate.
(first triangle set: figure with triangle xyz and triangle znm, right angles at y and n. options: not similar or not necessarily similar; similar: △xyz ~ △blank by the select.
second triangle set: triangle edf with sides 5, 7, 9 and triangle abc with sides 10, 14, 18. options: not similar or not necessarily similar; similar: △abc ~ △blank by the select.
third triangle set: triangle pqr with sides 12, 16 and included angle at q; triangle stu with sides 15, 20 and included angle at t. options: not similar or not necessarily similar; similar: △stu ~ △blank by the select.
explanation check)

Explanation:

Step1: Analyze First Triangle Pair

- Observe the first set of triangles: $\triangle XYZ$ and $\triangle MNZ$ (assuming the right - angled triangle with right angle at $Y$ and $N$). $\angle Y=\angle N = 90^{\circ}$, and $\angle XZY=\angle MZN$ (vertical angles are equal). By the AA (Angle - Angle) similarity criterion, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar. So $\triangle XYZ\sim\triangle MNZ$ by AA similarity.

Step2: Analyze Second Triangle Pair

- For $\triangle DEF$ (with sides $5,7,9$) and $\triangle ABC$ (with sides $10,14,18$). Check the ratios of corresponding sides: $\frac{5}{10}=\frac{1}{2}$, $\frac{7}{14}=\frac{1}{2}$, $\frac{9}{18}=\frac{1}{2}$. Since the ratios of all three corresponding sides are equal, by the SSS (Side - Side - Side) similarity criterion, $\triangle DEF\sim\triangle ABC$? Wait, no, the labels: $\triangle DEF$ has sides $DE = 5$, $DF=7$, $EF = 9$ and $\triangle ABC$ has $AB = 14$, $BC = 10$, $AC = 18$. Wait, $\frac{DE}{BC}=\frac{5}{10}=\frac{1}{2}$, $\frac{DF}{AB}=\frac{7}{14}=\frac{1}{2}$, $\frac{EF}{AC}=\frac{9}{18}=\frac{1}{2}$. So the correspondence is $\triangle DEF\sim\triangle BCA$? Wait, maybe I mislabeled. But the ratio of sides: $\frac{5}{10}=\frac{7}{14}=\frac{9}{18}=\frac{1}{2}$. So by SSS similarity, the triangles are similar. But wait, the option for the second pair: let's check the side lengths again. The small triangle has sides $5,7,9$, the large one has $10,14,18$. So $\frac{5}{10}=\frac{1}{2}$, $\frac{7}{14}=\frac{1}{2}$, $\frac{9}{18}=\frac{1}{2}$. So SSS similarity. But the option says "Not similar or not necessarily similar" or "Similar". Wait, maybe I made a mistake. Wait, the sides of the first triangle (small) are $5,7,9$ and the second (large) are $10,14,18$. So the ratios are equal. So they should be similar by SSS. But maybe the labels are different. Wait, the problem's second triangle: $\triangle ABC$ with $AC = 18$, $AB = 14$, $BC = 10$ and $\triangle DEF$ with $DE = 5$, $DF = 7$, $EF=9$. So $\frac{DE}{BC}=\frac{5}{10}=\frac{1}{2}$, $\frac{DF}{AB}=\frac{7}{14}=\frac{1}{2}$, $\frac{EF}{AC}=\frac{9}{18}=\frac{1}{2}$. So $\triangle DEF\sim\triangle BCA$ by SSS. But the option for the second pair: the radio button for "Not similar or not necessarily similar" or "Similar". Wait, maybe I miscalculated. Wait, $5/10 = 1/2$, $7/14=1/2$, $9/18 = 1/2$. So they are similar by SSS. But the problem's second pair: maybe the user made a typo, but let's proceed.

Step3: Analyze Third Triangle Pair

- For $\triangle PQR$ and $\triangle STU$. $\angle Q=\angle T$ (given as equal angles, vertical or marked equal). The sides around $\angle Q$: $PQ = 16$, $QR=12$ and around $\angle T$: $ST = 15$, $TU = 20$. Wait, no, $\frac{PQ}{TU}=\frac{16}{20}=\frac{4}{5}$, $\frac{QR}{ST}=\frac{12}{15}=\frac{4}{5}$. Since $\angle Q=\angle T$ and the sides around the angle are in proportion ($\frac{PQ}{TU}=\frac{QR}{ST}$), by the SAS (Side - Angle - Side) similarity criterion, $\triangle PQR\sim\triangle TUS$? Wait, the triangle is $\triangle STU$ and $\triangle PQR$. $\angle T=\angle Q$, $ST = 15$, $TU = 20$, $PQ = 16$, $QR = 12$. $\frac{ST}{QR}=\frac{15}{12}=\frac{5}{4}$, $\frac{TU}{PQ}=\frac{20}{16}=\frac{5}{4}$. So $\triangle STU\sim\triangle RQP$? Wait, maybe the correct correspondence is $\triangle STU\sim\triangle RQP$ by SAS? Wait, no, let's re - check. $\angle T$ and $\angle Q$ are equal. The sides adjacent to $\angle T$: $ST = 15$, $TU = 20$ and adjacent to $\angle Q$: $PQ = 16$, $QR = 12$. Wait, $\frac{ST}{QR}=\frac{15}{12}=\frac{5}{4}$, $\frac{TU}{PQ}=\frac{20}{16}=\frac{5}{4}$. So by SAS similarity, $\triangle STU\sim\triangle RQP$?

But let's go back to the first triangle:

First Triangle:
- $\triangle XYZ$ and $\triangle MNZ$: $\angle Y=\angle N = 90^{\circ}$, $\angle XZY=\angle MZN$ (vertical angles). So AA similarity. So $\triangle XYZ\sim\triangle MNZ$ by AA.

Second Triangle:
- Sides: $5,7,9$ and $10,14,18$. Ratios: $\frac{5}{10}=\frac{1}{2}$, $\frac{7}{14}=\frac{1}{2}$, $\frac{9}{18}=\frac{1}{2}$. So SSS similarity. So $\triangle DEF\sim\triangle ABC$ (if the correspondence is correct) by SSS.

Third Triangle:
- $\angle Q=\angle T$, $\frac{PQ}{TU}=\frac{16}{20}=\frac{4}{5}$, $\frac{QR}{ST}=\frac{12}{15}=\frac{4}{5}$. So SAS similarity. So $\triangle PQR\sim\triangle TUS$ (or $\triangle STU\sim\triangle RQP$) by SAS.

Answer:

1. First Pair: $\triangle XYZ\sim\triangle MNZ$ by AA (Angle - Angle) similarity.
2. Second Pair: $\triangle DEF\sim\triangle ABC$ by SSS (Side - Side - Side) similarity (since $\frac{5}{10}=\frac{7}{14}=\frac{9}{18}=\frac{1}{2}$).
3. Third Pair: $\triangle STU\sim\triangle RQP$ by SAS (Side - Angle - Side) similarity (since $\angle T=\angle Q$ and $\frac{ST}{QR}=\frac{TU}{PQ}=\frac{5}{4}$).

(Note: The answer is based on the standard triangle similarity criteria: AA (Angle - Angle), SSS (Side - Side - Side), SAS (Side - Angle - Side). The first triangle pair uses AA, the second SSS, and the third SAS.)