Question

are these triangles similar?
(image of two triangles: △qsr with angle at s (red) and angle at r (36°); △fgh with angle at h (36°) and angle at g (red))
options: yes (selected), no
write a similarity statement.
△qsr ~ △fgh

Explanation:

Step1: Check Angle Correspondence

In $\triangle QSR$, $\angle R = 36^\circ$ and $\angle S$ is marked (let's assume $\angle F$ and $\angle Q$ are right angles or we use AA similarity). In $\triangle FGH$, $\angle G = 36^\circ$ and $\angle H = 36^\circ$? Wait, no, $\angle H = 36^\circ$? Wait, $\triangle FGH$: $\angle H = 36^\circ$, $\angle G$ is marked, and $\angle F$ is a right angle? Wait, actually, $\angle S$ in $\triangle QSR$ and $\angle G$ in $\triangle FGH$ are marked (red angles), and $\angle R = 36^\circ$, $\angle H = 36^\circ$. So by AA (Angle - Angle) similarity, if two angles are equal, triangles are similar. $\angle R = \angle H = 36^\circ$, and the marked angles (let's say $\angle S = \angle G$). So AA similarity holds.

Step2: Write Similarity Statement

We need to match the angles. $\angle S$ corresponds to $\angle G$, $\angle R$ corresponds to $\angle H$, so $\triangle QSR \sim \triangle FGH$? Wait, no, wait: $\triangle QSR$: vertices Q, S, R. $\triangle FGH$: F, G, H. Wait, $\angle S$ (at S) and $\angle G$ (at G) are equal (marked), $\angle R = 36^\circ$ and $\angle H = 36^\circ$. So the correspondence is S - G, R - H, Q - F. So $\triangle QSR \sim \triangle FGH$? Wait, the given answer in the box is $\triangle QSR \sim \triangle FGH$, but let's check again. Wait, maybe $\triangle QSR \sim \triangle GFH$? No, the original problem has $\triangle QSR \sim \triangle FGH$ as filled. Wait, maybe the correct correspondence is $\triangle QSR \sim \triangle GFH$? No, the user's filled boxes are $\triangle QSR$ and $\triangle FGH$. Wait, let's confirm AA: two angles equal. $\angle R = 36^\circ$, $\angle H = 36^\circ$; and the other marked angle (at S and G) are equal. So by AA, triangles are similar, so the similarity statement is $\triangle QSR \sim \triangle FGH$ (or maybe $\triangle SRQ \sim \triangle GHF$? But the given answer is $\triangle QSR \sim \triangle FGH$).

Answer:

$\triangle QSR \sim \triangle FGH$ (and the first answer is "yes")