三角形和一点 仿射不变量

hbghlyj

反纯几何
import geometry;

unitsize(2 cm);

pair A = (-1,0);
pair B = (1,0);
pair C = (.6,1.5);

draw(A--B--C--cycle);

pair P = (0.3,0.6);
label("$P$", P, NE);

pair D = extension(A,P,B,C);
pair E = extension(B,P,C,A);
pair F = extension(C,P,A,B);

draw(A--D--F--E--D);
draw(A--F--C);
draw(B--E);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, -W);
label("$E$", E, NW);
label("$F$", F, S);

任意$\triangle ABC$和一点$P$, 证明$\frac{\cot \angle \mathrm{BAD}-\cot \angle \mathrm{ADE}}{\cot \angle \mathrm{CAD}-\cot \angle \mathrm{ADF}}$是仿射不变量.

hbghlyj

$DE\cap AB=X,\cot\angle BAD-\cot\angle ADE=\frac{AD}{d(X,AD)}$
$DF\cap AC=Y,\cot\angle CAD-\cot\angle ADF=\frac{AD}{d(Y,AD)}$

$d(X,AD)\over d(Y,AD)$是仿射不变量

hbghlyj

任意$\triangle A_1B_1C_1$和一点$P_1$,
任意$\triangle A_2B_2C_2$和一点$P_2$,
若$$\frac{\cot \angle B_1A_1D_1-\cot \angle  A_1D_1E_1}{\cot \angle C_1A_1D_1-\cot \angle A_1D_1F_1}=\frac{\cot \angle B_2A_2D_2-\cot \angle  A_2D_2E_2}{\cot \angle C_2A_2D_2-\cot \angle A_2D_2F_2}$$是否存在仿射变换 $f$ 使得 $f(A_1)=A_2,f(B_1)=B_2,f(C_1)=C_2,f(P_1)=P_2$