Question

question 9 of 10
is $\triangle bse \sim \triangle tes$? if so, identify the similarity postulate or
theorem that applies.

image of a geometric figure with right angles, segments labeled 5, 3, 5

a. similar - aa
b. similar - sss
c. similar - sas
d. cannot be determined

Explanation:

Step1: Analyze triangle sides and angles

In $\triangle BSE$ and $\triangle TES$:
- $BS = 5$, $ET = 5$ (given side lengths).
- $SE = 3$ (common side, or corresponding side).
- $\angle BSE$ and $\angle TES$ are right angles (since there are right angle symbols), so $\angle BSE=\angle TES = 90^\circ$.
- Also, we can check the ratio of sides: $\frac{BS}{TE}=\frac{5}{5} = 1$, $\frac{SE}{SE}=1$ (wait, no, let's correct. Wait, in $\triangle BSE$, sides are $BS = 5$, $SE = 3$, and in $\triangle TES$, sides are $TE = 5$, $SE = 3$, and the included angle between $BS$ and $SE$ is $\angle BSE$ (right angle), and between $TE$ and $SE$ is $\angle TES$ (right angle). So the ratio of two sides: $\frac{BS}{TE}=\frac{5}{5}=1$, $\frac{SE}{SE} = 1$? No, wait, maybe better to see SAS: two sides in proportion and included angle equal. Wait, $BS = TE = 5$, $SE$ is common, and $\angle BSE=\angle TES = 90^\circ$. So the ratio of $BS$ to $TE$ is $1$, ratio of $SE$ to $SE$ is $1$, and included angle (right angle) is equal. Wait, but actually, let's check the sides: in $\triangle BSE$, $BS = 5$, $SE = 3$; in $\triangle TES$, $TE = 5$, $SE = 3$. So $\frac{BS}{TE}=\frac{5}{5}=1$, $\frac{SE}{SE}=1$? No, that's not right. Wait, maybe the triangles are $\triangle BSE$ and $\triangle TES$. Let's label the angles: $\angle BSE$ is right angle, $\angle TES$ is right angle. Then, $BS = 5$, $ET = 5$; $SE = 3$ (common side). So the sides around the right angle: in $\triangle BSE$, legs are $BS = 5$ and $SE = 3$; in $\triangle TES$, legs are $TE = 5$ and $SE = 3$. So the ratio of the legs: $\frac{BS}{TE}=\frac{5}{5}=1$, $\frac{SE}{SE}=1$? No, that's not the right way. Wait, SAS similarity: if two sides are in proportion and the included angle is equal. Here, $BS = TE = 5$, $SE$ is common, and the included angle (right angle) is equal. Wait, but actually, the sides: $BS$ corresponds to $TE$, $SE$ corresponds to $SE$, and the included angle (right angle) is equal. So the ratio of $BS$ to $TE$ is $1$, ratio of $SE$ to $SE$ is $1$, and included angle is equal. But that would be SAS? Wait, no, SAS similarity requires two sides in proportion and included angle equal. Here, the sides are equal (ratio 1), so it's SAS similarity because two sides are in proportion (ratio 1) and included angle (right angle) is equal. Alternatively, AA: if two angles are equal. We know one right angle, and maybe another angle? Wait, in the parallelogram? Wait, the figure looks like a parallelogram, so $BT$ and $ES$ are parallel? Wait, maybe I made a mistake. Wait, let's re - examine. The triangles are $\triangle BSE$ and $\triangle TES$. $\angle BSE=\angle TES = 90^\circ$ (right angles). Now, $BS = 5$, $ET = 5$; $SE = 3$ (common side). So the sides: $BS/TE = 5/5 = 1$, $SE/SE = 1$? No, that's not the correct correspondence. Wait, maybe the correspondence is $B - T$, $S - E$, $E - S$? No, that's confusing. Wait, another approach: in $\triangle BSE$ and $\triangle TES$:
- $\angle BSE=\angle TES = 90^\circ$ (given right angles).
- $BS = ET = 5$ (given lengths).
- $SE$ is common. Wait, no, $SE$ is a side in both triangles. Wait, maybe the triangles are congruent, but the question is about similarity. But congruent triangles are similar with ratio 1. But let's check the similarity postulates. SAS similarity: two sides in proportion, included angle equal. Here, $BS/TE = 5/5 = 1$, $SE/SE = 1$, and included angle (right angle) is equal. So by SAS similarity, the triangles are similar. Alternatively, AA: if two angles are equal. We have one right angle, and maybe the other angles? Wait, in the figure, since $BS = ET = 5$ and $SE$ is common, and right angles, maybe the triangles are similar by SAS. Wait, but let's think again. The SAS similarity theorem states that if two sides of one triangle are proportional to two sides of another triangle and the included angles are congruent, then the triangles are similar. Here, $BS = TE = 5$, $SE$ is common (so $SE/SE = 1$), and the included angle (the right angle) is congruent. So the ratio of $BS$ to $TE$ is $1$, ratio of $SE$ to $SE$ is $1$, and included angle is equal. So by SAS similarity, the triangles are similar. Wait, but also, AA: if two angles are equal. We know one right angle, and maybe the other angles? Since $BS = TE$ and $SE$ is common, maybe the triangles are isoceles? No, maybe not. Wait, the correct answer is C: Similar - SAS. Because we have two sides in proportion (ratio 1) and included angle (right angle) equal.

Step2: Determine the similarity postulate

We have two sides ($BS$ and $SE$ in $\triangle BSE$; $TE$ and $SE$ in $\triangle TES$) with $BS = TE = 5$ and $SE$ is common (so the ratio of $BS$ to $TE$ is $1$ and the ratio of $SE$ to $SE$ is $1$), and the included angle (the right angle $\angle BSE$ and $\angle TES$) is equal. So by the SAS (Side - Angle - Side) similarity postulate, the triangles $\triangle BSE$ and $\triangle TES$ are similar.

Answer:

C. Similar - SAS