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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. |
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