Two triangles of equal area

Question

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?

Answer 1

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 $HGA$.

Answer 2

$I$ 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.

Answer 3

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

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, $|\triangle HGD|=|\triangle HGB|$.

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

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, $|\triangle HAG|=|\triangle HCG|$.