Here is the problem:
In the convex quadrilateral $ABCD$, it is known that $AD > BC$, points $E$ and $F$ are the midpoints of the diagonals $AC$ and $BD$, respectively, $EF = \frac{1}{2}(AD - BC)$. Prove that AD ∥ BC.
I thought that I'd proven it, but after a short review I understood that my proof is not correct. That's the short beginning that I think might lead to the whole solution:
- Let $K$ be the midpoint of the segment $AB$, draw the $KF$ line segment, $KF \cap AC = X$. So $KF$ is the midline of the $\triangle ABD$.
- Since $(EF = \frac{1}{2}(AD - BC))$ => $(AD = BC + 2EF)$ => $(KF = KX + XF = \frac{1}{2}BC + EF)$.
And now I'm stunned a bit, because intuitively it's clear that the points $E$ and $X$ are both the midpoints of the segment $AC$, which would help to prove what's asked, but I'm not sure how to prove it. How to do that? Alternative proofs are very welcome.
Please, note that this problem is only from an 8th grade school textbook and the answer must be within the given topic, which is simply "Midline of a triangle".
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