Question

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Explanation:

Step1: Analyze Triangle Similarity for Problem 7

First, identify the angles. $\angle Q$ and $\angle F$ are marked equal (given by the angle marks). $\angle QEB$ and $\angle FEG$ are vertical angles, so they are equal ($\angle QEB \cong \angle FEG$). 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, $\triangle QEB \sim \triangle FEG$? Wait, no, the triangle is $\triangle EFG$ and we need to find its similar triangle. Wait, the first triangle is $\triangle QBE$ (with vertices Q, B, E) and $\triangle FGE$? Wait, let's re - examine. $\angle Q=\angle F$ (marked), $\angle QEB=\angle FEG$ (vertical angles). So $\triangle QBE \sim \triangle FGE$? Wait, the problem says $\triangle EFG \sim$? Wait, maybe I misread. Let's look at the labels: Q, B, E and E, F, G. $\angle Q=\angle F$, $\angle QEB=\angle FEG$ (vertical angles). So by AA similarity, $\triangle QBE \sim \triangle FGE$? Wait, the triangle to compare with $\triangle EFG$: Wait, maybe the first triangle is $\triangle QBE$ and the second is $\triangle FGE$. Wait, perhaps the correct similar triangle for $\triangle EFG$ is $\triangle QBE$? Wait, no, let's check the vertices. $\triangle EFG$ has vertices E, F, G. $\triangle QBE$ has vertices Q, B, E. $\angle Q=\angle F$, $\angle QEB=\angle FEG$ (vertical angles), so AA similarity implies $\triangle QBE \sim \triangle FGE$? Wait, maybe the answer is $\triangle QBE$.

Step2: Analyze Triangle Similarity for Problem 8

For the second triangle, $\triangle GFE$ and we need to find its similar triangle. The lines $MN$ and $FE$ are parallel (marked by the same number of arrows, indicating parallel lines). So, $\angle F$ is common to both $\triangle GFE$ and $\triangle NME$? Wait, no, the segments $FM$ and $MN$? Wait, the lines $MN$ is parallel to $FE$ (since the arrows indicate parallelism). So, $\angle F=\angle NMG$ (corresponding angles, because $MN\parallel FE$ and $FG$ is a transversal), and $\angle G$ is common to both $\triangle GFE$ and $\triangle GMN$? Wait, no, let's see. $\triangle GFE$ and $\triangle GMN$: $\angle F=\angle NMG$ (corresponding angles, $MN\parallel FE$), $\angle G=\angle G$ (common angle). By AA similarity, $\triangle GFE \sim \triangle GMN$.

Answer:

For problem 7: $\triangle EFG \sim \triangle QBE$ (by AA similarity, $\angle Q = \angle F$, $\angle QEB=\angle FEG$)
For problem 8: $\triangle GFE \sim \triangle GMN$ (by AA similarity, $\angle F=\angle NMG$ (corresponding angles, $MN\parallel FE$), $\angle G=\angle G$ (common angle))