HELME PLEASE I WILL MARK THE BRAINEST
1 answer:
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Answer: AAA similarity.
Step-by-step explanation: CB is the transversal for the parallel lines AB and DE, and so by transverse property, we have ∠CED ≅ ∠CBA. Similarly, CA acts as a tranversal for the same pair of parallel lines AB and DE and using the same property, we can have ∠CDE ≅ ∠CAB. Now, in triangles CED and ABC, we have
∠CED ≅ ∠CBA,
∠CDE ≅ ∠CAB
and
∠DCE ≅ ∠ACB [same angle]
Hence, by AAA (angle-angle-angle) similarity,
△CED ~ △ABC.
Thus, the correct option is AAA similarity.
Answer:
Step-by-step explanation:
Standard form is basically where the degrees of the terms are in descending order.
First, let's distribute the negative sign to the terms in the second parentheses:
Now, we can just combine like terms with the terms from the first parentheses:
So, our answer is: