QUESTION IMAGE
Question
x y z n m not similar or not necessarily similar similar: △xyz ~ △ by the select a c 12 e b 6 6 3 d 4 f not similar or not necessarily similar similar: △abc ~ △ by the select
First Triangle Pair (△XYZ and △MNZ)
Step 1: Identify Right Angles
In $\triangle XYZ$, $\angle Y = 90^{\circ}$ (right angle), and in $\triangle MNZ$, $\angle N=90^{\circ}$ (right angle). So, $\angle Y=\angle N$.
Step 2: Identify Vertical Angles
$\angle XZY$ and $\angle MZN$ are vertical angles. Vertical angles are equal, so $\angle XZY = \angle MZN$.
Step 3: Apply AA Similarity
Since two angles of $\triangle XYZ$ are equal to two angles of $\triangle MNZ$ ($\angle Y=\angle N$ and $\angle XZY=\angle MZN$), by the AA (Angle - Angle) similarity criterion, $\triangle XYZ \sim \triangle MNZ$.
Second Triangle Pair (△ABC and △EFD - Check Proportions and Angle)
First, let's check the side lengths:
- In $\triangle ABC$: $AC = 12$, $BC = 6$
- In $\triangle EFD$: $EF=6$, $DF = 4$, $ED = 3$
Wait, maybe we made a mistake. Let's re - evaluate. Let's check the angles. Wait, maybe the correct triangle is $\triangle EFD$? Wait, no. Let's check the ratios.
Wait, in $\triangle ABC$, if we consider sides: Let's assume $\angle B=\angle D$? No, maybe we should check the ratios.
Wait, $AC = 12$, $BC = 6$, so $\frac{AC}{BC}=\frac{12}{6} = 2$. In $\triangle EFD$, $EF = 6$, $DF=4$, $ED = 3$. Wait, $\frac{EF}{ED}=\frac{6}{3}=2$, $\frac{DF}{?}$ Wait, maybe the angle between the sides.
Wait, maybe $\triangle ABC$ and $\triangle EFD$: Let's check the angles. Wait, maybe it's not similar? Wait, no. Wait, the first triangle pair is similar by AA, the second:
Wait, in $\triangle ABC$, sides $AC = 12$, $BC = 6$, and in $\triangle EFD$, sides $EF = 6$, $ED = 3$. The ratio of $AC$ to $EF$ is $\frac{12}{6}=2$, and the ratio of $BC$ to $ED$ is $\frac{6}{3} = 2$. If the included angle is equal, then by SAS similarity. But do we know the included angle?
Wait, maybe the problem has a typo, but for the first triangle:
The first triangle pair: $\triangle XYZ$ and $\triangle MNZ$ are similar by AA.
For the second triangle: Let's check the side lengths again. In $\triangle ABC$, $AC = 12$, $BC = 6$, $AB$ is unknown. In $\triangle EFD$, $EF = 6$, $ED = 3$, $DF = 4$. The ratio of $AC$ to $EF$ is $2:1$, the ratio of $BC$ to $ED$ is $2:1$. If $\angle C=\angle F$, then by SAS similarity. But since the problem says "similar", maybe $\triangle ABC\sim\triangle EFD$ by SAS (if the included angle is equal) or maybe it's a mistake. But based on the first part, the first triangle pair is similar by AA with $\triangle MNZ$.
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Similar: $\triangle XYZ \sim \triangle MNZ$ by the AA (Angle - Angle) Similarity Criterion