First we introduce notations: Problem. Given a quadrilateral ABCD with area S. Denote by A', B', C', D' the reflections of the points A, B, C, D in the lines BD, AC, BD, AC, respectively. Let S' be the area of the quadrilateral A'B'C'D'. Prove that $\frac{S^{\prime}}{S}<3$. In order to solve this problem, we switch to using directed areas (else, we could get in troubles with non-convex quadrilaterals). We will denote by $\left[P_1P_2...P_n\right]$ the directed area of a polygon $P_1P_2...P_n$, and, in particular, by $T=\left[ABCD\right]$ and $T^{\prime}=\left[A^{\prime}B^{\prime}C^{\prime}D^{\prime}\right]$ the directed areas of the quadrilaterals ABCD and A'B'C'D'. Then, the non-directed areas S and S' of these two quadrilaterals are the absolute values of these directed areas: S = |T| and S' = |T'|. Let the diagonals AC and BD of the quadrilateral ABCD intersect each other at the point F. It is almost obvious that the diagonals A'C' and B'D' of the quadrilateral A'B'C'D' intersect at F, too, but we won't need this in the rest of the proof . Instead, let's denote by t = < AFB the directed angle < AFB; hereby, "directed angle" means a directed angle modulo 360° (namely, this directed angle < AFB is the angle between the vectors $\overrightarrow{FA}$ and $\overrightarrow{FB}$) and not a directed angle modulo 180°. Then, < BFC = 180° ­- t, < CFD = t and < DFA = 180° - t. Since sin (180° - t) = sin t, we then have sin < AFB = sin < BFC = sin < CFD = sin < DFA = sin t. Since A' is the reflection of A in the line BD, we have FA' = FA and < BFA' = < AFB. Since B' is the reflection of B in the line AC, we have FB' = FB and < B'FA = < AFB. Thus, < B'FA' = < B'FA + < AFB + < BFA' = < AFB + < AFB + < AFB = t + t + t = 3t, and < A'FB' = - < B'FA' = - 3t. Now, since the area of a triangle equals $\frac12$ $\cdot$ the product of two sides $\cdot$ the sine of the angle between these sides, we have $\left[FAB\right]=\frac12\cdot FA\cdot FB\cdot\sin\measuredangle AFB$; $\left[FA^{\prime}B^{\prime}\right]=\frac12\cdot FA^{\prime}\cdot FB^{\prime}\cdot\sin\measuredangle A^{\prime}FB^{\prime}$. Since FA' = FA and FB' = FB, we thus have $\frac{\left[FA^{\prime}B^{\prime}\right]}{\left[FAB\right]}=\frac{\sin\measuredangle A^{\prime}FB^{\prime}}{\sin\measuredangle AFB}=\frac{\sin\left(-3t\right)}{\sin t}=\frac{-\sin\left(3t\right)}{\sin t}$, so that $\left[FA^{\prime}B^{\prime}\right]=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FAB\right]$. Similarly, $\left[FB^{\prime}C^{\prime}\right]=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FBC\right]$, $\left[FC^{\prime}D^{\prime}\right]=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FCD\right]$, $\left[FD^{\prime}A^{\prime}\right]=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FDA\right]$. Thus, $T^{\prime}=\left[A^{\prime}B^{\prime}C^{\prime}D^{\prime}\right]=\left[FA^{\prime}B^{\prime}\right]+\left[FB^{\prime}C^{\prime}\right]+\left[FC^{\prime}D^{\prime}\right]+\left[FD^{\prime}A^{\prime}\right]$ $=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FAB\right]+\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FBC\right]+\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FCD\right]+\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[FDA\right]$ $=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left(\left[FAB\right]+\left[FBC\right]+\left[FCD\right]+\left[FDA\right]\right)=\frac{-\sin\left(3t\right)}{\sin t}\cdot\left[ABCD\right]$ $=\frac{-\sin\left(3t\right)}{\sin t}\cdot T$, and $\frac{T^{\prime}}{T}=\frac{-\sin\left(3t\right)}{\sin t}$, so that $\frac{S^{\prime}}{S}=\frac{\left|T^{\prime}\right|}{\left|T\right|}=\left|\frac{T^{\prime}}{T}\right|=\left|\frac{-\sin\left(3t\right)}{\sin t}\right|=\left|\frac{\sin\left(3t\right)}{\sin t}\right|$. We have to prove that $\frac{S^{\prime}}{S}<3$; by the above, this is equivalent to $\left|\frac{\sin\left(3t\right)}{\sin t}\right|<3$. But this inequality is easy to prove: $\left\vert \frac{\sin \left( 3t\right) }{\sin t}\right\vert =\left\vert\frac{\sin \left( 2t+t\right) }{\sin t}\right\vert =\left\vert \frac{\sin\left( 2t\right) \cos t+\cos \left( 2t\right) \sin t}{\sin t}\right\vert=\left\vert \frac{\sin \left( 2t\right) \cos t}{\sin t}+\cos \left(2t\right) \right\vert $ $=\left\vert \frac{2\sin t\cos t\cos t}{\sin t}+\cos \left( 2t\right)\right\vert =\left\vert 2\cos ^{2}t+\cos \left( 2t\right) \right\vert \leq\left\vert 2\cos ^{2}t\right\vert +\left\vert \cos \left( 2t\right)\right\vert<3$, since $\left|2\cos^2 t\right|\leq 2$ and $\left|\cos\left(2t\right)\right|\leq 1$ with equality only for t being a multiple of 180°, what is not possible in our case since t is the angle < AFB between the diagonals AC and BD of the (non-degenerate) quadrilateral ABCD. Proof complete. Darij

Я сечас умру от тоски. Let $\alpha$ be the acute angle betwen the diagonals, then after the reflection the angle between diagonals becomes $3\alpha$ or $3\alpha-\pi$, moreover the length of diagonals are preserved. It follows that \[\frac{S'}{S}=\left|\frac{\sin 3\alpha}{\sin\alpha}\right|=|3-4\sin^2\alpha|<3.\]