Question

given that \\(\frac{ab}{de} = \frac{bc}{ef} = \frac{1}{2}\\), complete the statements to show that \\(\triangle abc \sim \triangle def\\) by the sas similarity theorem. horizontal and vertical lines are perpendicular. so, angles \\(\boldsymbol{\text{dropdown}}\\) are right angles by definition of perpendicular. all right angles \\(\boldsymbol{\text{dropdown}}\\) congruent. therefore, \\(\boldsymbol{\text{dropdown}}\\) def by the sas similarity theorem. the image shows a coordinate grid with triangles abc and def. points: a(-2,7), b(-2,3), c(-6,3); d(-2,-10), e(-2,-2), f(6,-2).

Explanation:

Step1: Analyze Triangle ABC

In $\triangle ABC$, points $B(-2,3)$ and $C(-6,3)$ have the same $y$-coordinate, so $BC$ is horizontal. Points $A(-2,7)$ and $B(-2,3)$ have the same $x$-coordinate, so $AB$ is vertical. Thus, $\angle B$ is the right angle (between horizontal $BC$ and vertical $AB$).

Step2: Analyze Triangle DEF

In $\triangle DEF$, points $E(-2,-2)$ and $F(6,-2)$ have the same $y$-coordinate, so $EF$ is horizontal. Points $D(-2,-10)$ and $E(-2,-2)$ have the same $x$-coordinate, so $DE$ is vertical. Thus, $\angle E$ is the right angle (between horizontal $EF$ and vertical $DE$).

Step3: Identify Congruent Angles

We need the included angle between the sides with the given ratio. For $\triangle ABC \sim \triangle DEF$ by SAS similarity, the included angles must be congruent. The sides $AB$ (vertical in $\triangle ABC$) and $DE$ (vertical in $\triangle DEF$), and $BC$ (horizontal in $\triangle ABC$) and $EF$ (horizontal in $\triangle DEF$) form angles $\angle B$ (in $\triangle ABC$) and $\angle E$ (in $\triangle DEF$), which are both right angles. So the angles are $\angle B$ and $\angle E$.

Answer:

B and E