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How can I prove that if two triangles which form by continuing the sides of a convex quadrilateral until intersection have equal areas, then one of the diagonals of a quadrilateral bisects the other?

Drawing

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Since area $ABE$ = area $CBF$, we must also have area $ACE$ = area $ACF$. Then altitudes of $E$ and $F$ from $AC$ are the same, and $EF$ is parallel to $AC$.

Now $\triangle ABC$ is similar to $\triangle BEF$ so $B$ lies on the line joining the midpoints of $AC$ and $EF$: join the $B$ to the midpoint of the base for each of $\triangle ABC$ and $\triangle BEF$, and check the matching angles to see that the two segments form a straight line

Because $\triangle ACD$ is similar to $\triangle DEF$, the line from D to the midpoint of $EF$ also passes through the midpoint of $AC$ thus also through $B$.

Thus $BD$ bisects $AC$.

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    Now △ABC is similar to △BEF so BB lies on the line joining the midpoints of AС and EF How is this theorem called? – 2017-01-20
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    I don't know if it is a named theorem. To see that it is so, join the apex $B$ to the midpoint of the base for each of $\triangle ABC$ and $\triangle BEF$, and check the angles to see that the two segments form a straight line. – 2017-01-20
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By proving that $AC$ is parallel to $EF$, which can be deduced from

either the fact that

$$AB \cdot BE \cdot \sin(\beta) = Area(ABE) = Area(CBF) = CB \cdot BF \cdot \sin(\beta)$$ where $\angle \, ABE = \angle \, CBF = \beta$,

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from the fact that $Area(ABE) = Area(CBF)$ implies $Area(AFE) = Area(CFE)$, which in its own turn leads to the conclusion that the altitude from vertex $A$ of triangle $AFE$ is equal (and obviously parallel) to the altitude from vertex $C$ of triangle $CFE$.