A geometry problem about areas and lengths. Squares and polygons.

Question

An area inequality:
Let $ ABCD $ be a square, $ AB = 1 $. Let's consider a 4-sided polygon $ EFGH $, with a peak on each side of the square. The area of EFGH equals $ \frac{1}{2} $. Prove that it exists a line intersecting two of $ EFGH $'s sides, paralell to one of the polygon's sides, with it's length greater or equal then $ \frac{1}{2} $.
I have proved that one of the $ EG $, $ FH $ is parallel to the square's sides, using a geometrical area spliting -> the interior polygon made by the intersections of parallels from each of the $ E, F, G, H $ to the square's sides have area 0.

Answer 1

Consider square to have diagonal from $(0,0)$ to $(1,1)$. Let's denote with $L(x')$ the length of the cross-section of $EFGH$ with line $x=x'$.

If we assume that $\forall x: L(x)<\frac12$, then the area of $EFGH$: $$ A_{EFHG} = \int_0^1L(x)dx<\int_0^1\frac12dx=\frac12, $$ which is false by the problem setting. Thus $\exists x:L(x)\ge\frac12$

Edit: By the way, we never used the fact that $EFGH$ is a quadrilateral. So the result is true for any integrable shape that is anchored at opposite sides of a square and has simple-connected parallel cross-sections.