Proofs

Yeah, the third side is the base and is half the length of it. Look at it here!

Isn't that the Midpoint segment thrm? Yeah, it is! So, the midsegment of a triangle is parallel and is half the length of the third side by the Triangle Midsegment Theorem.

No, it isn't half the length of the third side, it's half the length of the base, and parallel to the third side. I don't know that theorem, that's why I need help. Thanks for trying, but whenever anyone other than CPhill helps me, it never works.

;)))))))))))))

Shades....most of this is straight out of Euclid's Elements......Book VI....Proposition 2

Here it is: http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI2.html

I won't repeat it verbatim, because I can't do a better job than Eulclid....LOL!!!!

You need to use the second part  under the long dashed line that begins with...

"Next, let the sides AB and AC of the triangle ABC be cut proportionally, so that BD is to AD as CE is to AE. Join DE."

At the end of the proof....it is shown that the lines are parallel.....now....all we need to add is this :

m<ADE  = m<ABC       the corresponding angles formed by a transversal  [in this case, the transversal AB]  cutting two parallel lines are congruent

Thus, by AA congruency.....triangle ADE ≈ triangle ABC

And.....similar triangles are similar in all respects......thus AD  is half of AB....therefore....DE  is half of BC  [ thus....the line segment joining the sides  = 1/2 of the base]

Let me know if you get stuck......I'll see if I can help you out....!!!!