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Newton's theorem and related corollaries

2022-08-02 01:33:00 【patrickpdx】

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 $A BC D$ There is an inscribed circle, cut separately $A B$ , $BC$ , $C D$ , $D A$ 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. $EG$ , $F H$ , $B D$ 交于一点, 设该点为 $M$ , 且 $BM / D M = BE / HD$ .

分别设 $EG$ , $F H$ 交 $B D$ 于点 $M_{1}$ , $M_{2}$ . $\frac{BM_{1}}{DM_{1}}=\frac{S_{\triangle BEM_{1}}}{S_{\triangle DGM_{1}}}\cdot \frac{GM_{1}}{EM_{1}}$ . 由于 $\angle AEG=\angle DGE$ , $\frac{S_{\triangle BEM_{1}}}{S_{\triangle DGM_{1}}}=\frac{EM_{1}\cdot BE}{GM_{1}\cdot DG}$ , 故 $\frac{BM_{1}}{DM_{1}}=\frac{BE}{EG}=\frac{BE}{HD}$ . 同理可证: $\frac{BM_{2}}{DM_{2}}=\frac{BE}{HD}$ . 证毕.

根据引理, 可以推得 $EG$ , $F H$ , $A C$ Also hand in one point, 命题得证.

When quadrilateral $A BC D$ When it is a concave quadrilateral, The proof is exactly the same.

When quadrilateral $A BC D$ 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 $S_{\triangle MID}=S_{\triangle MIB}$

设 $S_{\triangle AID}=S_{1}$ , $S_{\triangle CID}=S_{2}$ , $S_{\triangle BIC}=S_{3}$ , $S_{\triangle AIB}=S_{4}$

$S_{\triangle MID}=S_{2}-\frac{1}{2}S_{\triangle ACD}-\frac{1}{2}S_{\triangle IAC}=S_{2}-\frac{1}{2}S_{\triangle ACD}-\frac{1}{2}(S_{1}+S_{2}-S_{\triangle ACD})=\frac{1}{2}(S_{2}-S_{1})$