Showing that two triangle have the same content Euclidean style proof. 
We know that D is the midpoint of AB and that E is the midpoint of AC. 
We wish to show that Triangle DBC has the came content as Triangle BEC. if we can do so we have a theorem that will tell us that DE is parallel to the line BC.
This question is asked in the spirit of euclid. no midpoint theorem or proving the midpoint theorem in anther way is allowed.
Edit: Could we perhaps argue that Triangle DEC and Triangle ADE have the same base and Height somehow?
 A: SKETCH OF PROOF
Find two ways to 'shear' triangle $\Delta{CDE}$ into $\Delta{BDE}$. One of these shearing transformations is 'indirect' and keeps the content of the triangle constant at all times, ensuring that $\Delta{CDE}$ = $\Delta{ADE}$ = $\Delta{BDE}$ in content. Such equality of content means the second shearing action, being 'direct', must be constrained by two parallel lines. Those parallel lines are $\overleftrightarrow{BC}$ and $\overleftrightarrow{DE}$, Q.E.D. While shearing is strictly speaking not a Euclidean concept, the Euclidean propositions I.36 to I.39 required for this proof can be expressed in the modern language of shearing transformations. Note as well that this rather modern logic based on the 'shearing' of equal-content figures is actually very basic in Euclid's proof system. It appears in Elements I, while SAS similarity must await his sixth (!) volume.
DETAILS OF PROOF


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*Euclid's Proposition I.36, expressed in modern language, says that a parallelogram keeps its area, i.e., its 'content', even as it is sheared by sliding its 'top' and 'bottom' bases relative to each other along their 'baselines'.

*Viewing a triangle as a degenerate parallelogram whose 'top' side has zero length (but Euclid may have had problems understanding the concept of zero, forcing him to prove I.37 first), we get I.38: a triangle keeps its content even as its 'top' vertex and 'bottom' base slide relative to each other along their 'baselines'.

*Fix 'top' vertex $D$ in place, but slide 'bottom' base $\overline{CE}$ along $\overleftrightarrow{AC}$ until the base coincides with $\overline{AE}$. By I.38, $\Delta{CDE}$ and $\Delta{ADE}$ have equal content. 

*Then, view $\Delta{ADE}$ as having 'top' vertex $E$ and 'bottom' base $\overline{AD}$. Fix $E$ in place, but slide 'bottom' base $\overline{AD}$ along $\overleftrightarrow{AB}$ until the base coincides with $\overline{BD}$. By I.38 again, $\Delta{ADE}$ and $\Delta{BDE}$ have equal content. 

*By the transitivity of equality (a.k.a. 'Common Notion 1' in Euclid's time), $\Delta{ADE}$, $\Delta{BDE}$, and $\Delta{CDE}$ have equal content. More specifically, $\Delta{BDE}$ = $\Delta{CDE}$ in content.

*I.39 is the converse of I.37: A shearing transformation that maintains the area of a triangle whose 'bottom' base is fixed must slide the 'top' vertex along a line parallel to the fixed 'bottom' baseline. 

*By I.39, the shearing transformation defined by the equal-content triangles $\Delta{BDE}$ and $\Delta{CDE}$, whose fixed 'bottom' base is $\overline{BC}$, must be a shearing action wherein $B$ slides to $C$ along a line $\overleftrightarrow{BC}$ that is parallel to the 'bottom' baseline $\overleftrightarrow{DE}$, which was what we wanted.
NOTES


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*First, strictly speaking, Euclid did not view I.36 to I.39 as statements on shearing transformations. While Euclid defined congruence in terms of superposition, which is one kind of transformation, mathematicians viewed transformations suspiciously until the 19th Century. 

*Two tools Euclid used to prove I.36 to I.38 are adding and subtracting content, i.e., cut-and-paste. He, however, neither explicitly axiomized nor axiomatically proved that content is a quantity amenable to cut-and-paste.

*Euclid also used reductio ad absurdum, i.e., proof by contradiction, to prove I.39. Some mathematicians do not accept proof by contradiction, though, given that apparent proofs of contradiction stalled the rise of non-Euclidean geometries for nearly two millennia.

*Note as well that while Euclid's logic based on the 'shearing' of equal-content figures is surprisingly compatible with modern transformation-based geometry, it is actually near-foundational in his proof system. All the Euclidean propositions used above appear in Elements I, while the SAS similarity argument most contemporary students would employ in this proof had to await Euclid's sixth (!) volume. $\blacksquare$
