Consider a circular labelling, A, B, C, D, of the vertices of the quadrangle, and let the diagonals $AC$ and $BD$ meet at $O$. Without loss of generality, we may (and will) assume that the line $\ell$ meets the opposite sides $AB$ and $CD$, say, at $X$ and $Y$, respectively. Further, let $\alpha = \angle AOX = \angle COY$ and $\beta = \angle BOX = \angle DOY$, write $\text{area}[AOB] = \text{area}[AOX] + \text{area}[BOX]$ and express

$\text{area}[AOB] = \frac{1}{2} \cdot OA \cdot OB \cdot \sin(\alpha + \beta)$, $\text{area}[AOX] = \frac{1}{2} \cdot OA \cdot OX \cdot \sin \alpha$ and $\text{area}[BOX] = \frac{1}{2} \cdot OB \cdot OX \cdot \sin \beta$, to get

$$\begin{align*} OX &= \frac{OA \cdot OB \cdot \sin(\alpha + \beta)}{OA \cdot \sin \alpha + OB \cdot \sin \beta} \le \frac{OA \cdot OB \cdot (\sin \alpha + \sin \beta)}{OA \cdot \sin \alpha + OB \cdot \sin \beta} \\ &\le \frac{OA \cdot \sin \beta + OB \cdot \sin \alpha}{\sin \alpha + \sin \beta}. \end{align*}$$

(The last inequality is easily seen to be equivalent to the obvious inequality $(OA - OB)^2 \cdot \sin \alpha \cdot \sin \beta \ge 0$.) Similarly, $OY \le (OC \cdot \sin \beta + OD \cdot \sin \alpha)/(\sin \alpha + \sin \beta)$, so $XY \le (AC \cdot \sin \beta + BD \cdot \sin \alpha)/(\sin \alpha + \sin \beta) \le \max(AC, BD)$.

Let $\mathcal{L}$ be the pencil of lines through the point of intersection of the diagonals of $K$. We shall prove that the length of each segment of intersection $\ell \cap K$, $\ell \in \mathcal{L}$, does not exceed the length of the longest diagonal of $K$.

To begin, notice that no intersection segment has a length greater than the diameter of $K$, so $\sup_{\ell \in \mathcal{L}} |\ell \cap K|$ is finite, where $|s|$ denotes the length of the line-segment $s$.

Suppose, if possible, that $\sup_{\ell \in \mathcal{L}} |\ell \cap K|$ is greater than the length of the longest diagonal of $K$. Let $\delta$ and $\delta'$ denote the diagonals of $K$ and consider a line $\ell_0$ in $\mathcal{L}$ such that

$$|\ell_0 \cap K| > \frac{\sup_{\ell \in \mathcal{L}} |\ell \cap K| + \max(|\delta|, |\delta'|)}{2} > \frac{|\delta| + |\delta'|}{2}.$$

Recall that the length of the internal bisectrix of an angle of a triangle is smaller than the arithmetic mean of the lengths of the sides forming that angle, to infer that $\ell_0$ does not bisect the corresponding angle formed by the diagonals; say, the angle formed by $\ell_0$ and $\delta$ is smaller than the angle formed by $\ell_0$ and $\delta'$, both angles being, of course, those in the wedge containing $\ell_0$.

Finally, reflect the line of support of $\delta$ in the line $\ell_0$ to obtain a line $\ell_1$ in $\mathcal{L}$ such that

$$\frac{|\ell_1 \cap K| + \max(|\delta|, |\delta'|)}{2} \ge \frac{|\ell_1 \cap K| + |\delta|}{2} > |\ell_0 \cap K| > \frac{\sup_{\ell \in \mathcal{L}} |\ell \cap K| + \max(|\delta|, |\delta'|)}{2},$$

and thereby reach a contradiction. (The inequality in the middle expresses the above mentioned fact about the length of an internal bisectrix in a triangle.) The conclusion follows.