Reciprocal of midsegment (midpoint) theorem proof I need a hint on how to prove that "In a triangle ABC if D is the midpoint of the side AB and the line through D intersects AC at E such that BC=2DE then E is the midpoint of AC and BC is parallel to DE"
Thank you
 A: Edit:
a proof of the question as stated does not exist, as it's easy to construct a counterexample:
A circle of radius ½BC centred on D intersects (the infinite continuation of) AC at E and E’; for some triangles E’ will be between A and C, and therefore a line DE’ has length ½BC while not being parallel to BC.
E’ is a solution to "intersects at E" in the problem statement, which implies a "for all E" quantifier. If you're missing a "there exists" quantifier for E, then that quantifier would also allow satisfaction of "two-sides-and-any-angle makes a similar triangle", rebutting your objection in the comments below.
Original:
Looks like a simple case of similar triangles, therefore the angles are the same, therefore...
Start by constructing DE as parallel, then show that it's half of BC. Then work backwards to form a complete proof.
Remember to confirm that parallellism is both necessary and sufficient (ie the interference is reversible).
A: In $\triangle ABC$ , if we apply the law of sine, we get:
$$ \frac{BC}{\sin\angle BAC} = \frac{AB}{\sin\angle ACB}$$
or, $$\frac{BC}{AB} = \frac{\sin\angle BAC}{\sin\angle ACB}    $$
Now, In $\triangle ADE$
$$ \frac{DE}{\sin\angle BAC} = \frac{AD}{\sin\angle AED}$$
or, $$ \frac{\frac{BC}{2}}{\sin\angle BAC} = \frac{\frac{AB}{2}}{\sin\angle AED}$$
or, $$\frac{BC}{AB} = \frac{\sin\angle BAC}{\sin\angle AED} $$
Therefore, $$\sin\angle ACB = \sin\angle AED$$
A: Name the midpoint of BC as F and join this point with the midpoint of AB (i.e. D). With this, use the midpoint theorem to conclude that DF is parallel to AC. With a little angle chasing and using one of the congruence theorems, we can prove that $\Delta ADE$ is congruent to $\Delta DBF$. From here, it will be obvious that DE is parallel to BC.
