Question:
On a coordinate plane, â–³ABC has points (-2, 7), (-2, 3), and (-6, 3) and â–³DEF has points (-2,-10), (-2, -2), and (6, -2).
Given that
[tex]\frac{AB}{DE} = \frac{BC}{EF} = \frac{1}{2}[/tex]
Complete the statements to show that â–³ABC ~ â–³DEF by the SAS similarity theorem.
Horizontal and vertical lines are __(1)__. So, angles are right angles by definition of perpendicular lines. All right angles are __(2)__
Therefore, â–³ABC ~ â–³DEF by the SAS similarity theorem.
Answer:
(1) perpendicular
(2) congruent
Step-by-step explanation:
Required
Fill in the gaps
[tex]\frac{AB}{DE} = \frac{BC}{EF} = \frac{1}{2}[/tex] implies that:
[tex]AB : DE= BC : EF[/tex]
In other words:
AB ~ DE ---- Side (S)
and
BC ~ EF ---- Side (S)
Now to fill in the gap with:
(1) perpendicular
(2) congruent
Further explanation
(1) Vertical lines and horizontal lines meet at a right angle (i.e. 90 degrees). Any two lines that meet at a right angle are perpendicular
(2) Angles with equal measures are congruent. Because right angles have a congruent angles of 90 degrees, then they are congruent.
Answer:
Given that StartFraction A B Over D E EndFraction = StartFraction B C Over E F EndFraction = one-half, complete the statements to show that â–³ABC ~ â–³DEF by the SAS similarity theorem.
Horizontal and vertical lines are
✔ perpendicular
.
So, angles
✔ B and E
are right angles by definition of perpendicular lines.
All right angles are
✔ congruent
.
Therefore, â–³ABC ~ â–³DEF by the SAS similarity theorem.
Step-by-step explanation:
