Question

The vertices of the triangles are located as follows: upper A at left parenthesis negative 4 comma negative 6 right parenthesis, upper B at left parenthesis 3 comma negative 6 right parenthesis, upper C at left parenthesis negative 2 comma negative 1 right parenthesis, upper A prime at left parenthesis negative 4 comma negative 3 right parenthesis, upper B prime at left parenthesis 10 comma negative 3 right parenthesis, and upper C prime at left parenthesis 0 comma 7 right parenthesis.

How can a similarity transformation be used to determine that the AA criterion proves the dilated triangle A′B′C′
𝐴
′
𝐵
′
𝐶
′
is similar to triangle ABC
𝐴
𝐵
𝐶
?

Translate the dilated triangle so one of its angles exactly matches the corresponding angle in the original triangle; repeat using a different pair of angles.
Translate the dilated triangle so one of its angles exactly matches the corresponding angle in the original triangle; repeat using a different pair of angles.

Find the angle measures of the original triangle and the dilated triangle, then compare proportionality of the ratio to the scale factor.
Find the angle measures of the original triangle and the dilated triangle, then compare proportionality of the ratio to the scale factor.

Center the original triangle inside the dilated triangle and see if the side lengths are congruent by using the scale factor.
Center the original triangle inside the dilated triangle and see if the side lengths are congruent by using the scale factor.

Reflect the triangle across the x-axis and check for symmetry among the triangles with respect to the x-axis.

Answer 1

Use angles (AA). Show two pairs of corresponding sides are parallel so two pairs of corresponding angles are equal.

Compute slopes:
- AB: slope = (−6 − (−6))/(3 − (−4)) = 0. So AB is horizontal.
- A′B′: slope = (−3 − (−3))/(10 − (−4)) = 0. So A′B′ is also horizontal ⇒ ∠A = ∠A′ (the directions of the first side at A and A′ match).
- AC: slope = (−1 − (−6))/(−2 − (−4)) = 5/2.
- A′C′: slope = (7 − (−3))/(0 − (−4)) = 10/4 = 5/2 ⇒ AC ∥ A′C′ ⇒ the second angle at A equals the second angle at A′.

Thus two corresponding angles are equal, so by AA the triangles are similar.

(As extra confirmation: there is a dilation with center (−4,−9) and scale factor 2 taking A,B,C to A′,B′,C′, which also preserves angles and gives similarity.)