Let $ABCD$ be a convex quadrilateral with $AB = CD = 10$, $BC = 14$, and $AD = 2\sqrt{65}$. Assume that the diagonals of $ABCD$ intersect at point $P$, and that the sum of the areas of triangles $APB$ and $CPD$ equals the sum of the areas of triangles $BPC$ and $APD$. Find the area of quadrilateral $ABCD$.
Let $AP=x$ and let $PC=\rho x$. Let $[ABP]=\Delta$ and let $[ADP]=\Lambda$. We easily get $[PBC]=\rho \Delta$ and $[PCD]=\rho\Lambda$. 
We are given that $[ABP] +[PCD] = [PBC]+[ADP]$, which we can now write as \[\Delta + \rho\Lambda = \rho\Delta + \Lambda \qquad \Longrightarrow \qquad \Delta -\Lambda = \rho (\Delta -\Lambda).\] Either $\Delta = \Lambda$ or $\rho=1$. The former would imply that $ABCD$ is a parallelogram, which it isn't; therefore we conclude $\rho=1$ and $P$ is the midpoint of $AC$. Let $\angle BAD = \theta$ and $\angle BCD = \phi$. Then $[ABCD]=2\cdot [BCD]=140\sin\phi$. On one hand, since $[ABD]=[BCD]$, we have \begin{align}\sqrt{65}\sin\theta = 7\sin\phi \quad \implies \quad 16+49\cos^2\phi = 65\cos^2\theta\end{align}whereas, on the other hand, using cosine formula to get the length of $BD$, we get \[10^2+4\cdot 65 - 40\sqrt{65}\cos\theta = 10^2+14^2-280\cos\phi\]\begin{align}\tag{2}\implies \qquad  65\cos^2\theta = \left(7\cos\phi+ \frac{8}{5}\right)^2\end{align}Eliminating $\cos\theta$ in the above two equations and solving for $\cos\phi$ we get\[\cos\phi = \frac{3}{5}\qquad \implies \qquad \sin\phi = \frac{4}{5}\]which finally yields $[ABCD]=2\cdot [BCD] = 140\sin\phi = 112$.