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I'm interested in the following Euclidean geometry problem:

$ABCD$ is a cyclic quadrilateral.
$M$ is the midpoint of $CD$.
$N$ is such that $AMBN$ is a harmonic quadrilateral.
$E$ is the intersection of lines $AC$ and $BD$.
$F$ is the intersection of lines $AD$ and $BC$.
Prove that $E,F,N$ are collinear.

From https://web.evanchen.cc/handouts/cmplx/en-cmplx.pdf Lemma 8.

If $A, B, C, D$ lie on the unit circle then the intersection of $\overline{A B}$ and $\overline{C D}$ is given by $$ \frac{a b(c+d)-c d(a+b)}{a b-c d} . $$

Apply this formula to $E=AC\cap BD$ and $F=AD\cap BC$ $$ e=\frac{a c(b+d)-b d(a+c)}{a c-b d}\\ f=\frac{a d(b+c)-b c(a+d)}{a d-b c} $$

According to http://users.math.uoc.gr/~pamfilos/eGallery/problems/HarmonicQuad.html

A cyclic quadrangle $ABCD$ is called harmonic when (considering the points as complex numbers) the cross-ratio $(ABCD)=-1$.

$AMBN$ is a harmonic quadrilateral. So $(AMBN)=-1$. $$ (a - m) (b - n) + (a - n) (b - m)=0\\ m=\frac{c+d}2 $$ Solving for $n$, $$n=\frac{a (-4 b+c+d)+b (c+d)}{2 (a+b-c-d)}$$ To show that $E,F,N$ are collinear, I need to show that $\frac{e-n}{f-n}$ is real. $$\frac{e-n}{f-n}=\frac{a d-b c}{a c-b d}\cdot\frac{2 a b (c+d)-a c (c+3 d)-b d (3 c+d)+2 c d (c+d)}{2 a b (c+d)-a d (3 c+d)-b c (c+3 d)+2 c d (c+d)}$$ Since $A,B,C,D$ lie on the unit circle, $a^*=a^{-1},b^*=b^{-1},c^*=c^{-1},d^*=d^{-1}$, $$\left(\frac{e-n}{f-n}\right)^*=\frac{a^{-1} d^{-1}-b^{-1} c^{-1}}{a^{-1} c^{-1}-b^{-1} d^{-1}}\cdot\frac{2 a^{-1} b^{-1} (c^{-1}+d^{-1})-a^{-1} c^{-1} (c^{-1}+3 d^{-1})-b^{-1} d^{-1} (3 c^{-1}+d^{-1})+2 c^{-1} d^{-1} (c^{-1}+d^{-1})}{2 a^{-1} b^{-1} (c^{-1}+d^{-1})-a^{-1} d^{-1} (3 c^{-1}+d^{-1})-b^{-1} c^{-1} (c^{-1}+3 d^{-1})+2 c^{-1} d^{-1} (c^{-1}+d^{-1})}$$ multiply both the numerator and denominator of the first fraction by $abcd$, multiply both the numerator and denominator of the second fraction by $abc^2d^2$, to get $\left(\frac{e-n}{f-n}\right)^*=\frac{e-n}{f-n}$, so $\frac{e-n}{f-n}$ is real.

Is there a better way to prove that $E,F,N$ are collinear?

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$% Mathematica code: % Block[{a=Exp[110Degree I],b=Exp[70Degree I],c=Exp[-20Degree I],d=(-a b+2 a c-b c)/(a-2 b+c),m=(c+d)/2,e=(a c(b+d)-b d(a+c))/(a c-b d),f=(a d(b+c)-b c(a+d))/(a d-b c),n=(2 a b-a m-b m)/(a+b-2 m)},Show[Graphics[Circle[]],Graphics[Line[ReIm/@{d,c,f,d,b,a,c}]],ComplexListPlot[{Callout[a,"a",Above],Callout[b,"b"],Callout[c,"c"],Callout[d,"d",Before],Callout[e,"e",Above],Callout[f,"f"],Callout[m,"m"],Callout[n,"n"]}],Axes->False]]$

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To show that $\frac{e - n}{f - n}$ is real, it suffices to prove that $\frac{e - n}{f - n} = \overline{\left(\frac{e - n}{f - n}\right)}$ using the formula of $\frac{e - n}{f - n}$ involving only $a, b, c, d$. This is quite doable because for every $w$ on the unit circle, $\bar{w} = \frac{1}{w}$.

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

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.

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