Let ${P}_{1}$ and ${P}_{2}$ be two convex quadrilaterals such that ${P}_{1}\ne {P}_{2}$ and $Area({P}_{1})\ge Area({P}_{2})$. Is it true that it is not possible that all sides and diagonals of ${P}_{1}$ are shorter than the corresponding sides and diagonals of ${P}_{2}$?

Suppose that all four sides and one diagonal of ${P}_{1}$ are shorter than the corresponding sides and diagonal of ${P}_{2}$. Then, as it is explained in the answer to my older question, the two triangles which triangulate ${P}_{2}$ should be "flatter" (otherwise it's not possible $Area({P}_{1})\ge Area({P}_{2})$). From this it should follow that the other diagonal of ${P}_{1}$ is longer than the corresponding diagonal of ${P}_{2}$.

Do you think it's correct? Can you help me formalizing it?

Suppose that all four sides and one diagonal of ${P}_{1}$ are shorter than the corresponding sides and diagonal of ${P}_{2}$. Then, as it is explained in the answer to my older question, the two triangles which triangulate ${P}_{2}$ should be "flatter" (otherwise it's not possible $Area({P}_{1})\ge Area({P}_{2})$). From this it should follow that the other diagonal of ${P}_{1}$ is longer than the corresponding diagonal of ${P}_{2}$.

Do you think it's correct? Can you help me formalizing it?