Given convex $\square ABCD$, and $H\in (AB)$ and $G\in (AD)$ such that $(CH)\parallel(AD)$ and $(CG)\parallel(AB)$, define $E=(BG)\cap(HD)$. Prove that: $$S_{AHEG}=S_{HBCD}$$ (where $S$ denotes area).
My try:
We know that: $AHCG$ is parallelogram. For example area $S_{AHCG}=2z$.Connect $[HG],[EC],[EA]$. Then $S_{AHE}+S_{ECG}=z$ also $S_{HCE}+S_{EGA}=z$
From here I don't know how I complete?