The
Triangle Midsegment Theorem:

ÒIn a
triangle, the segment joining the midpoints of any two sides will be parallel
to the third side and half its length.Ó

Consider
the triangle below:

Construct
a line through C that is parallel to AB.

We know
the following equalities by the midpoint construction:

AD = DC
and AE = EB. We can use the reflexive property of equality for angle A to
obtain a SAS similarity of triangles DAE and CAB. Now that the two triangles
are congruent, we have m<ADE = m<ACB. By the converse of the
Corresponding Angles Theorem, we have DE || CB.

Now we
need to show that DE = ½ CB.

Claim:
Quadrilateral CBEF is a parallelogram. If we can show FC=EB, we can conclude
that CBEF is a parallelogram.

*(side): We have m<FCD = m<EAD by the Alternate Interior
Angles Theorem

*(angle): We have AD = DC (from above – definition of
midpoint)

*(side): We have m<FDC = m<ADE by the Vertical Angles
Theorem.

**We now
have Triangle FDC is congruent to Triangle EDA by the SAS Congruence Theorem.

Therefore,
FC = AE --> FC = EB. If a pair of opposite sides of a
quadrilateral are both congruent and parallel, then the quadrilateral is
parallelogram (previous theorem).
Therefore FE = CB. Also, the above triangle congruence gives us FD = DE.

Write: FE
= FD + DE --> FE = 2DE --> DE = ½ FE

Since FE
= CB, we can substitute: DE = ½ CB.

Therefore
the Midsegment is half of the measure of the third side.

Conclusion:
If the midpoints of any two sides of a triangle are joined by
a segment, then that segment is parallel to the third side and half its
length.