Answer:
A) DC equals EC times AC all over BC.
Step-by-step explanation:
We know that:
[tex]DE\parallel AB[/tex]
Then by corresponding angles:
[tex]\angle CAB\cong \angle CDE\text{ and } \angle CBA\cong \angle CED[/tex]
So, by AA-Similarity:
[tex]\Delta CDE\sim \Delta CAB[/tex]
Corresponding sides of similar triangles are in proportion. Hence:
[tex]\displaystyle \frac{DC}{AC}=\frac{EC}{BC}[/tex]
By multiplying both sides by AC:
[tex]\displaystyle DC=\frac{EC\cdot AC}{BC}[/tex]
DC is equal to EC times AC over BC.
Hence, our answer is A.