In convex quadrilateral $ABCD$, $AB=2$, $AD=4$, and $2BC+CD=10$. If angle $DAC$ equals angle $DBC$, and the diagonals of $ABCD$ are perpindicular to each other, what is the area of $ABCD$?
I have a solution but it is too ugly and definitley not the intended one. Due to my incompetence in geometry, I don't see the simple solution.
Edit nvm I see the answer. I posted without thinking too hard. Sorry!
By Ptolemy, $2(10-a)+4a=(x+w)(y+z)$ where we break up diagonal $AC$ into segments with lengths $x,w$ and similarly for the other diagonal into segments $y,z$, at the intersection point of the diagonals. So $20=xy+xz+wy+wz=2A$ by breaking up into right triangles with vertex at diagonal intersection point, so $A=10$.
I realize this is poorly written but no time sorry.