Newton's theorem and related corollaries

摘自 《高中数学竞赛解题策略·几何分册》Newton's theorem, See full text illustrations link

Newton's theorem. 四边形 $ABCD$ There is an inscribed circle, cut separately $AB$, $BC$, $CD$, $DA$ 于点 $E$, $F$, $G$, $H$. Then the intersection of the diagonals of the quadrilateral and the line connecting the tangent points of the two opposite sides intersect at the same point.

证明: We first prove the case of convex quadrilaterals.

引理. $EG$, $FH$, $BD$ 交于一点, 设该点为 $M$, 且 $BM/DM=BE/HD$.

分别设 $EG$, $FH$ 交 $BD$ 于点 $M_1$, $M_2$. $\frac{B{M}_1}{D{M}_1}=\frac{S_{△BE{M}_1}}{S_{△DG{M}_1}}\cdot \frac{G{M}_1}{E{M}_1}$. 由于 $\angle AEG=\angle DGE$, $\frac{S_{△BE{M}_1}}{S_{△DG{M}_1}}=\frac{E{M}_1\cdot BE}{G{M}_1\cdot DG}$, 故 $\frac{B{M}_1}{D{M}_1}=\frac{BE}{EG}=\frac{BE}{HD}$. 同理可证: $\frac{B{M}_2}{D{M}_2}=\frac{BE}{HD}$. 证毕.

根据引理, 可以推得 $EG$, $FH$, $AC$ Also hand in one point, 命题得证.

When quadrilateral $ABCD$ When it is a concave quadrilateral, The proof is exactly the same.

When quadrilateral $ABCD$ When it is a folded quadrilateral, The proof is exactly the same.

性质1. The midpoints of the two diagonals of the inscribed quadrilateral are collinear with the center of the inscribed circle.

证明:

如图, 证明 $S_{△MID}=S_{△MIB}$

设 $S_{△AID}=S_1$, $S_{△CID}=S_2$, $S_{△BIC}=S_3$, $S_{△AIB}=S_4$

$S_{△MID}=S_2-\frac{1}{2}{S}_{△ACD}-\frac{1}{2}{S}_{△IAC}=S_2-\frac{1}{2}{S}_{△ACD}-\frac{1}{2}(S_1+S_2-S_{△ACD})=\frac{1}{2}(S_2-S_1)$

同理 $S_{△MIB}=\frac{1}{2}(S_3-S_4)$

易证, $(S_2-S_1)=\frac{1}{2}r(CD-AC)$, $(S_3-S_4)=\frac{1}{2}r(BC-AB)$, 显然相等, 证毕.

证明参考 link