Triangle Geometry

Medians AD and BE of triangle ABC are perpendicular. If AD = 15 and BE = 20,

then what is the area of triangle ABC?

not true to scale

$\begin{array}{|rcll|} \hline A_{\triangle ABC}=A &=& \dfrac{ab\sin{C}}{2} \\ \hline \\ A_{\triangle EDC} &=& \dfrac{\dfrac{ab\sin{C}}{4}}{2} \\\\ &=& \dfrac{\dfrac{ab\sin{C}}{2}}{4} \\\\ &=& \dfrac{A}{4} \\ \hline \\ A_{ABDE} &=& \dfrac{AD\cdot BE}{2} \\\\ &=& \dfrac{15\cdot 20}{2} \\\\ &=& 15\cdot 10 \\\\ &=& 150 \\ \hline \\ A &=& A_{\triangle EDC} + A_{ABDE} \\ A &=& \dfrac{A}{4} + 150 \quad & | \quad \cdot 4 \\ 4A &=& A + 600 \\ 3A &=& 600 \quad & | \quad :3 \\ \mathbf{A} &\mathbf{=}& \mathbf{200} \\ \hline \end{array}$

The area of triangle ABC is 200

CPPhill: I think heureka's reasoning comes from this property of irregular quadrilateral:

"The line joining the mid-points of any two adjacent sides is parallel to the corresponding diagonal and equal to half of it."