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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 $Area(\triangle HGC)=Area(\triangle HAG)$.

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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?

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    $\begingroup$ I would say that the proof is ok 👌 $\endgroup$ – Dr. Mathva Jan 13 at 9:50
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    $\begingroup$ Setting aside the result in question ... Anne's Theorem "works" for any point on the Newton line, except that for points outside the quadrilateral, we have that the sum of two areas equals the difference of the other two. Points on the boundary of the quadrilateral are the literal edge case where sum-vs-difference doesn't matter because one of the areas is zero. $\endgroup$ – Blue Jan 13 at 9:50
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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 $HGA$.

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Drop the perpendicular from $D$ onto $\overline{GH}$ at $E$.
Drop the perpendicular from $B$ onto $\overline{GH}$ at $F$.

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Since $I$ bisects $\overline{BD}$, we have $\overline{ID}=\overline{IB}$. Furthermore, $\angle DIE=\angle BIF$ since they are vertically opposite. Thus, $\triangle DIE=\triangle BIF$. Therefore, the altitude $\overline{DE}$ is the same as the altitude $\overline{BF}$ and thus, $|\triangle HGD|=|\triangle HGB|$.


Drop the perpendicular from $C$ onto $\overline{GH}$ at $E$.
Drop the perpendicular from $A$ onto $\overline{GH}$ at $F$.

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Since $J$ bisects $\overline{AC}$, we have $\overline{JA}=\overline{JC}$. Furthermore, $\angle AJH=\angle CJG$ since they are vertically opposite. Thus, $\triangle AFJ=\triangle CEJ$. Therefore, the altitude $\overline{AF}$ is the same as the altitude $\overline{CE}$ and thus, $|\triangle HAG|=|\triangle HCG|$.

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$I$ is the midpoint of diagonal $BD$ by definition.

So $\triangle HBI$ and $\triangle HDI$ have the same area (same base length along $BD$ and same height from $H$ onto $BD$), and $\triangle GBI$ and $\triangle GDI$ have the same area.

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