Claim. Given a convex quadrilateral $ABCD$ that isn't parallelogram. The Newton line intersects sides $AD$ and $BC$ of the quadrilateral at the points $H$ and $G$ , respectively . Then $Area(\triangle HGD)=Area(\triangle HBG)$ and $Area(\triangle HGC)=Area(\triangle HAG)$.
GeoGebra applet that demonstrates this claim can be found here.
Proof. Observe that diagonal $HG$ of the quadrilateral $HBGD$ lie on the Newton line of the quadrilateral $HBGD$. If we apply Anne's theorem on point $H$ and quadrilateral $HBGD$ we can write $Area(\triangle HGD)+0=Area(\triangle HBG)+0$ , hence $Area(\triangle HGD)=Area(\triangle HBG)$. Similarlly we can show that $Area(\triangle HGC)=Area(\triangle HAG)$.
Question. Is this proof acceptable? Can we use Anne's theorem in the case when the point lies on the side of the quadrilateral? Can you provide an alternative proof?
Best Answer
Drop from $B$ and $D$ the altitudes $BB'$ and $DD'$ to common base $HG$: triangles $BB'I$ and $DD'I$ are congruent by $ASA$. Hence $BB'\cong DD'$ and $Area(HGD)=Area(HGB)$.
An analogous proof can be repeated for triangles $HGC$ and $HGA$.