When D and E are the midpoints of the sides AC and BC we get the following situation.
Since CD=AD and CE=EB, we have that AC = 2DC and BC=2EB. This allows us to write the following proportion.
AC/DC = BC/EC = 2 ∠C ≅ ∠C
The Side-Angle-Side (SAS) Similarity Theorem implies that â–³ ABC ~ â–³ DEC. Thus, we write the following equation.
2 = AC/DC = BC/EC = AB/DE
⇓
DE = 1/2 AB
The latter equation is the one that the Triangle Midsegment Theorem states. Consequently, the Triangle Midsegment Theorem is a specific case of the Triangle Proportionality Theorem when the segment parallel to a side passes through the midpoints of the other two sides.
Extra
Extra
Notice that we did not need DE to be parallel to AB to establish the similarity between △ ABC and △ DEC. In fact, by using this similarity, we get that ∠CDE ≅ ∠A and thus, the Corresponding Angles Converse implies that DE∥ AB.
