The two triangles C'A'B' and CAB are similar by the Side-Angle-Side (SAS) similarity criterion. The angles at A and A' are equal since they are both vertically opposite, and the lengths of the sides opposite these angles (C'A' and CA, B'A' and AB) are proportional, as the side lengths of C'A'B' (10 and 14) when divided by the respective side lengths of triangle CAB (5 and 7) yield consistent ratios. This satisfies the conditions for the SAS similarity criterion, confirming that the triangles are indeed similar.
A picture is shown of two triangles. The first labeled C'A'B', and the second labeled CAB. Corner C' of C'A'B' is located at (-6, 8), A' is located at (-6, -2), and B' is located at (8, -2). Meanwhile, corner C of triangle CAB is located at (-3, 3) while A is located at (-3, -2), and B is located at (4, -2).
Using any of the three triangle similarities of corresponding angles or side lengths, explain in 1–2 sentences why the two triangles are similar. Reference the triangle similarity criteria of choice (SAS, SSS, or AA) and explain your reasoning in your answer.
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Are B' and B also vertically opposite?
No, B' and B are not vertically opposite angles. Vertically opposite angles occur when two lines intersect, creating pairs of opposite angles that are equal. In the case of triangles C'A'B' and CAB, only A and A' are angles that correspond as vertical angles. To clarify the similarity of the triangles, we can still state that AA similarity is applicable here since they share a common angle (angle A and angle A') and have another pair of corresponding angles (angles B and B' and angles C and C'), which are equal due to the property of the triangle. Thus, the triangles C'A'B' and CAB are similar by the Angle-Angle (AA) similarity criterion.
1 answer