Question

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.

Answer 1

∠A=∠E (90°)

∠ACB=∠DCE(vertically opposite angle)

therefore ΔABC ~ ΔEDC (AA criteria)

Answer 2

The correct answer that can be chosen is

15/5 = 12/4= 9/3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate.

Given:

ΔABC and ΔEDC are two triangles

To find:

Prove that ΔABC and ΔEDC are similar.

Solution:

Condition used:

When two triangles are similar the ratio of corresponding angles will equal and the corresponding angles will be equal.

From the triangle ABC,

AB = 3, AC = 4, and BC = 5

From the triangle DEC

DE = 9, EC = 12, and DC = 15

By using the above condition,

The ratio of corresponding sides = AB/DE = AC/EC = BC/DC

=> 3/9 = 4/12 = 5/15

=> 1/3 = 1/3 = 1/3

Here the ratio of corresponding sides is equal

Therefore,

It is proven that triangles ΔABC and ΔEDC are similar to each other.

The correct answer that can be chosen is

15/5 = 12/4= 9/3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate.

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