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 dropdown. so, angles dropdown are right angles as a result of perpendicular lines. all right angles are dropdown. therefore, \\(\triangle abc \sim \triangle def\\) by the theorem. the dropdown options include congruent, parallel, perpendicular. the coordinates are: a(-2, 7), b(-2, 3), c(-6, 3), d(-2, -10), e(-2, -2), f(6, -2).

Explanation:

Step1: Analyze Horizontal/Vertical Lines

Horizontal and vertical lines are perpendicular (since horizontal has slope 0, vertical is undefined, product of slopes is -1, so they intersect at 90°). So the first blank: check angles \( \angle ABC \) and \( \angle DEF \). For \( \triangle ABC \), \( B(-2,3) \), \( A(-2,7) \) (vertical line), \( C(-6,3) \) (horizontal line), so \( \angle ABC \) is right. For \( \triangle DEF \), \( E(-2,-2) \), \( D(-2,-10) \) (vertical), \( F(6,-2) \) (horizontal), so \( \angle DEF \) is right. So horizontal/vertical lines are perpendicular, making those angles right angles.

Step2: Right Angles Property

All right angles are congruent (by definition, right angles are 90°, so they are equal in measure, hence congruent).

Step3: SAS Similarity

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

Answer:

First dropdown: \( \angle ABC \) and \( \angle DEF \) (but from the options' context, the first blank's line: "Horizontal and vertical lines are \(\boldsymbol{\text{perpendicular}}\)" (since they form right angles). Second blank: "All right angles are \(\boldsymbol{\text{congruent}}\)". So filling:
First: perpendicular
Second: congruent