Let $ABCD$ a convex quadrilateral such that $AB=CD$. Let $P$ and $Q$ are the mid points of the sides $BC$ and $AD$ respectively. Now if we joint $PQ$, is it divide the quadrilateral in equal area?
To show that, I used trapizium as a counter-example but it didn't work. Is it true? Or there is some quadrilateral which can disprove the statement. Please help me to solve this.
If it's true, so since $$S_{\Delta APQ}=S_{\Delta DPQ},$$ we obtain $$S_{\Delta ABP}=S_{\Delta DCP},$$ which gives $$\sin\measuredangle ABP=\sin\measuredangle DCP,$$ which is wrong in the general.
Just, take a quadrilateral $ABCD$ such that $AB=CD$ and $\measuredangle B\neq\measuredangle C$ and $\measuredangle B+\measuredangle C\neq180^{\circ}.$