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Prove that ΔABC and ΔEDC are similar.

Answer choices:
15 over 5 equals 12 over 4 equals 9 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate.

∠DCE is congruent to ∠CBA by the Vertical Angles Theorem and 15 over 5 equals 12 over 4 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate.

∠E and ∠B are right angles and, therefore, congruent since all right angles are congruent. 9 over 4 and 12 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Postulate.

15 over 4 equals 12 over 5 equals 9 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Postulate.​

Prove That ΔABC And ΔEDC Are SimilarAnswer Choices 15 Over 5 Equals 12 Over 4 Equals 9 Over 3 Shows The Corresponding Sides Are Proportional Therefore ΔABC ΔEDC class=

Sagot :

The ratio of the corresponding sides of the triangles is the same that is 3. Then ΔABC is similar to ΔEDC.

What is the triangle?

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

The triangles are given below.

ΔABC and ΔEDC

If the ratio of the corresponding sides of the triangles is the same. Then the triangles aid to be a similar triangle.

9 / 3 = 3

12 / 4 = 3

15 / 5 = 3

The ratio of the corresponding sides of the triangles is the same that is 3. Then the triangles aid to be a similar triangle.

If any two angles of the triangles are the same then the triangles are similar triangles.

∠A = ∠E = 90°

∠ACB = ∠ECD (vertically opposite angle)

More about the triangle link is given below.

https://brainly.com/question/25813512

#SPJ1

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