Question

Prove that in any triangle the line joining the mid-points of any two sides is parallel to the third side and half of its length.

Answer 1

Answer on Question #46180 – Math – Geometry

Prove that in any triangle the line joining the mid-points of any two sides is parallel to the third side and half of its length.

Proof

Let $AD = BD$ and $AE = CE$. Prove that $DE \mid |BC$ and $DE = BC/2$.

Extend DE beyond E to F such that $DE = EF$. Since $AE = CE$, triangles ADE and CEF are equal, making $CF \mid |AB$ (or $CF \mid |BD$, which is the same) because, for the transversal AC, the alternating angles DAE and ECF are equal. Also, $CF = AD = BD$, such that BDFC is a parallelogram. It follows that $BC = DF = 2 \cdot DE$ which is what we set out to prove.

Conversely, let D be on AB, E on AC, DE $\mid$ BC and DE = BC/2. Prove that AD = DB and AE = CE.

This is so because the condition $DE \mid |BC$ makes triangles ADE and ABC similar, with implied proportion,

$$AB/AD = AC/AE = BC/DE = 2.$$

It thus follows that AB is twice as long as AD so that D is the midpoint of AB; similarly, E is the midpoint of AC.

www.AssignmentExpert.com