几何画板自定义工具过4点的抛物线是什么原理

hbghlyj

以C为中心将CB旋转∠DCA,以A为中心将AB旋转∠DAC,两直线交于E,
圆O为ABC外接圆,N为过E作的圆O的一条切线上的动点
以C为中心将CB旋转∠NCA,以A为中心将AB旋转∠NAC,两直线交于F
则F的轨迹为过四点A,B,C,D的抛物线

hbghlyj

发到 math.stackexchange.com/questions/5016915/ 问问

hbghlyj

啊原来就是直线的等角共轭

@Intelligenti pauca
answered 7 mins ago

This construction follows from a nice property of isogonal conjugation:

The isogonal conjugate of a line with respect to triangle $ABC$ is a conic passing through $A$, $B$ and $C$; specifically, an ellipse, parabola, or hyperbola according as the line intersects the circumcircle of $ABC$ in $0$, $1$, or $2$ points.

Here's then how it works.

1. We construct the isogonal conjugate $E$ of point $D$ with respect to $ABC$.

2. As $D$ lies on the parabola we want to find, point $E$ must belong to its isogonal conjugate, i.e. a line tangent to the circumcircle of $ABC$.

3. We construct then one such tangent $EN$ and take an arbtrary point $N$ on it.

4. We construct the isogonal conjugate $F$ of point $N$ with respect to $ABC$. Point $F$ belongs then to the parabola.