Question

applying the sas similarity theorem
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 \\(\square\\).
so, angles \\(\square\\) are right angles by definition of perpendicular lines.
all right angles are \\(\square\\).
therefore, \\(\triangle abc \sim \square\\) by sas similarity theorem.
(options for the last dropdown: congruent, supplementary, vertical)
coordinates (from image): a(-2,7), b(-2,3), c(-6,3), e(-2,-2), f(6,-2), d(-2,-10)

Explanation:

Step1: Analyze horizontal and vertical lines

Horizontal and vertical lines are perpendicular (since horizontal lines have a slope of 0 and vertical lines have an undefined slope, and their product is -1, indicating perpendicularity).

Step2: Identify right angles

Looking at the coordinates: In $\triangle ABC$, $B(-2,3)$, $A(-2,7)$ (vertical line $AB$) and $B(-2,3)$, $C(-6,3)$ (horizontal line $BC$), so $\angle ABC$ is a right angle. In $\triangle DEF$, $E(-2,-2)$, $D(-2,-10)$ (vertical line $DE$) and $E(-2,-2)$, $F(6,-2)$ (horizontal line $EF$), so $\angle DEF$ is a right angle. So angles $\angle ABC$ and $\angle DEF$ are right angles.

Step3: Property of right angles

All right angles are congruent (by the definition of right angles, each is $90^\circ$, so they are congruent).

Step4: Apply SAS similarity

Given $\frac{AB}{DE}=\frac{BC}{EF}=\frac{1}{2}$ and the included angles (the right angles) are congruent, by SAS similarity theorem, $\triangle ABC \sim \triangle DEF$.

Answer:

1. Horizontal and vertical lines are \(\boldsymbol{\text{perpendicular}}\).
2. Angles \(\boldsymbol{\angle ABC}\) and \(\boldsymbol{\angle DEF}\) are right angles.
3. All right angles are \(\boldsymbol{\text{congruent}}\).
4. Therefore, \(\triangle ABC \sim \boldsymbol{\triangle DEF}\) by the SAS similarity theorem.