Let $A_1A_2\dots A_n$ be the given polygon. Take any $A_i,A_j,A_k,A_{\ell}$ and let $A_iA_k$ and $A_jA_{\ell}$ meet at $Q$. It boils down to show $A_iQ,A_jQ,A_kQ,A_{\ell}Q$ all have rational lengths. We know that all sides and diagonals of the convex quadrilateral $A_iA_jA_kA_{\ell}$ have rational lengths. Let $\angle A_kA_iA_j=\alpha$ and $\angle A_kA_iA_{\ell}=\beta$ (see the figure). Applying the law of cosines for the triangles $\triangle A_iA_jA_k, A_iA_kA_{\ell}, \triangle A_iA_jA_{\ell}$ we get that $\cos\alpha, \cos\beta,\cos(\alpha+\beta)$ are all rationals. It means that $\sin\alpha\sin\beta$ is also rational. Further, we have $$\frac{A_jQ}{QA_{\ell}}=\frac{S_{\triangle A_iQA_j}}{S_{\triangle A_iQA_{\ell}}}=\frac{A_iA_j\cdot \sin\alpha}{A_iA_{\ell}\cdot\sin\beta}$$Since $\frac{\sin\alpha}{\sin\beta}=\frac{\sin\alpha\sin\beta}{\sin^2\beta}$ and $\sin^2 \beta, \sin\alpha\sin\beta$ are both rationals, it follows $\frac{A_jQ}{QA_{\ell}}$ is rational, hence both segments $A_jQ,QA_{\ell}$ have rational lengths. With a similar argument we get $A_iQ,QA_k$ are rationals.

USAMO 2003 P2