调和四边形 证明共线

hbghlyj

ABCD 是一个圆内接四边形，M 是 CD 中点，点 N 满足 AMBN 是调和四边形，E 是直线 AC,BD 的交点，F 是直线 AD,BC 交点，证明 EFN 共线

出处:旧帖子的第1题.

hbghlyj

已经在 MSE 解决了

Here is another proof that doesn't use a coordinate system.


Let $J$ be the intersection of $AB$ and $CD$, $H$ be the intersection of $EF$ and $CD$, then $(J, H, D, C) = -1$. Because $M$ is the midpoint of $CD$ and $(J, H, D, C) = -1$, $\overline{JC}\cdot\overline{JD} = \overline{JH}\cdot\overline{JM}$. Moreover, $\overline{JA}\cdot\overline{JB} = \overline{JC}\cdot\overline{JD}$, so $\overline{JA}\cdot\overline{JB} = \overline{JH}\cdot\overline{JM}$, which means $A, B, M, H$ are concyclic.

On the other hand, $(HA, HB, HN, HM) = -1$ (because $AMBN$ is a harmonic quadrilateral), and $(HA, HB, HF, HM) = -1$, so the lines $HF$ and $HN$ are identical. Therefore $E, N, F$ are collinear.

hbghlyj

从1#的图中可以看出，该红线上还有A、B处切线的交点L，如何证明？