How prove that $AD$ and $BC$ of the given convex quadrilateral are parallel?

Question

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:

1. 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 $\triangle ABD$.
2. 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".

Answer 1

Let $K$ be the midpoint of $AB$, as in your attempt.

Consider $\triangle ABD$, then $KF = \frac12 AD$.

Consider $\triangle ABC$, then $KE = \frac12 BC$.

Verify that $KE + EF = KF$, so $KEF$ is a straight line.

Answer 2

You have the right idea. However, instead of introducing $X$, just draw from $E$ to the midpoint of $AB$, i.e., your $K$. Since by SAS we have $\triangle AEK \sim \triangle ACB$, we then get

$$\lvert KE\rvert = \frac{1}{2}\lvert BC\rvert \tag{1}\label{eq1A}$$

Similarly, $\triangle BKF \sim \triangle BAD$ so

$$\lvert KF\rvert = \frac{1}{2}\lvert AD\rvert \tag{2}\label{eq2A}$$

Using what's given, as well as \eqref{eq1A} and \eqref{eq2A}, we have

$$\lvert KE\rvert + \lvert EF\rvert = \frac{1}{2}\lvert BC\rvert + \frac{1}{2}(\lvert AD\rvert - \lvert BC\rvert) = \frac{1}{2}\lvert AD\rvert = \lvert KF\rvert \tag{3}\label{eq3A}$$

This shows that $KEF$ is a straight line. From the triangle similarity conditions (e.g., $\measuredangle AKE = \measuredangle ABC$ and $\measuredangle AEK = \measuredangle ACB$), we get $KE \parallel BC$ and $EF \parallel AD$. Thus, this means that

$$AD \parallel BC \tag{4}\label{eq4A}$$