The converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
From the sketch,
[tex]\frac{CD}{DA}=\frac{CE}{EB}[/tex]
Take the reciprocal
[tex]\frac{DA}{CD}=\frac{EB}{CE}[/tex]
[tex]\text{Add 1 TO BOTH SIDES}[/tex]
[tex]\frac{DA}{CD}+1=\frac{EB}{CE}+1[/tex]
Any number divided by itself is 1, so we can replace 1 with CD/CD or CE/CE
so that
[tex]\frac{CD}{CD}+\frac{DA}{CD}=\frac{CE}{CE}+\frac{EB}{CE}[/tex]
Combine terms using our common denominator
[tex]\frac{CD+DA}{CD}=\frac{CE+EB}{CE}[/tex]
from the diagram, we can see that
CA=CD+DA
and
CB=CE+EB
Then
[tex]\frac{CA}{CD}=\frac{CB}{CE}[/tex]
Since the triangles have SAS for triangle similarity.
This means that
[tex]\text{Triangle ABD is similar to Triangle }CDE[/tex]
