# Prove 4 points cyclic in an acute scalene triangle

$ABC$ is an acute angled scalene triangle. $L,M,N$ are midpoints of the sides $BC,CA,AB$.

The perpendicular bisectors of $\overline{AB}$ and $\overline{CA}$ meet $\overline{AL}$ at point $D$ and point $E$. The rays $\overrightarrow{BD}$ and $\overrightarrow{CE}$ cut each other at point $F$ inside the triangle.

Prove that $A,M,F,N$ are cyclic.

Tried by taking a point on symmedian and also tried by applying Menelaus theorem.

## 1 Answer

Let $O=ME \cap ND$. By construction $O$ is the center of circumscribed circle of $\triangle ABC$ with circumradius $R=|AO|$. Right triangles $ANO$ and $AOM$ share common hypotenuse $AO$, which serves as the diameter of the circumscribed circle with the center $O_c$ in the middle of $AO$ for both of them, and the radius of the circle is $\tfrac12|AO|=\tfrac12R$.

So, to prove that $A,M,F,N$ are cyclic, suffice it to show that $|O_cF|=|O_cA|$ or $|O_cF|^2=|O_cA|^2$.

Considering the points as complex numbers, let

\begin{align} P&=0,\quad A=-4w,\quad B=12w ,\\ C&=A+4(u+v\cdot i)=4(u-w+v\cdot i) ,\\ L&=\tfrac12\,(B+C)=2u+4w+2v\cdot i ,\\ M&=\tfrac12\,(A+C)=2u-4w+2v\cdot i ,\\ N&=\tfrac12\,(A+B)=4w ,\\ Q&=\tfrac12\,(A+M)=u-4w+v\cdot i \end{align}
for some real numbers $u,v,w$.

The lines through points $z_1,z_2$ and $z_3,z_4$ intersect at a point

\begin{align} z&= f_{\times}(z_1,z_2,z_3,z_4)= \frac{ (z_1-z_2)\,(\overline{z_3}\,z_4-\overline{z_4}\,z_3) -(z_3-z_4)\,(\overline{z_1}\,z_2-\overline{z_2}\,z_1) }{ (z_1-z_2)\,(\overline{z_3}-\overline{z_4}) -(z_3-z_4)\,(\overline{z_1}-\overline{z_2}) } , \end{align} where $\overline{z}$ is the complex conjugate of $z$.

The center $O_c$ can be found as the point of intersection of perpendicular bisectors of $AM$ and $AN$.

The perpendicular bisector of $AN$ is defined by points $P$ and $P+i$, the perpendicular bisector of $AM$ is defined by points $Q$ and $Q+(C-A)i$. Given that,

\begin{align} Q_c&=f_{\times}(P,P+i,Q,Q+(C-A)\cdot i) =\frac1v\,(v^2-4uw+u^2)\cdot i . \end{align}

The other points can be found similarly, \begin{align} D&=f_{\times}(N,N+i,A,L) =4w+\frac{8vw}{u+4w}\cdot i ,\\ E&= f_{\times}(M,M+O_c-Q,A,L) \\ &=2\cdot\frac{u(u+4w)(u-2w)+v^2(u+2w)}{u^2+4uw+v^2} +2\cdot\frac{v(u^2+v^2)}{u^2+4uw+v^2}\cdot i ,\\ F&= f_{\times}(C,E,B,D) \\ &= 4\cdot\frac{(8uw+3u^2+3v^2-16w^2)\,w}{(u+4w)^2+v^2} +64\cdot\frac{v w^2}{(u+4w)^2+v^2}\cdot i . \end{align}

Now it can be verifies that \begin{align} |O_cF|^2&=|O_cA|^2 =\frac1{v^2}(v^2+u^2)(v^2+(u-4w)^2) . \end{align}

The following maxima code can be used to verify the result:

_(z):=conjugate(z)$xpoint(z1,z2,z3,z4):= ((z1-z2)*(_(z3)*z4-_(z4)*z3)-(z3-z4)*(_(z1)*z2-_(z2)*z1)) /((z1-z2)*(_(z3)-_(z4))-(z3-z4)*(_(z1)-_(z2)))$
declare([P,A,B,C,L,M,N,Q,Oc,D,E,F],complex)$declare([u,v,w],real)$
P:0;
A:-4*w;
B:12*w;
C:A+4*(u+%i*v);
L:(B+C)/2;
M:(A+C)/2;
N:(A+B)/2;
Q:(A+M)/2;
Oc:factor(xpoint(P,P+%i,Q,Q+(C-A)*%i));
t:factor(xpoint(N,N+%i,A,L));
D:factor(realpart(t))+%i*imagpart(t);
t:factor(xpoint(M,M+Oc-Q,A,L));
E:realpart(t)+%i*imagpart(t);
t:factor(xpoint(C,E,B,D));
F:realpart(t)+%i*imagpart(t);
OcF2:factor(cabs(F-Oc)^2);
OcA2:factor(cabs(A-Oc)^2);
OcA2-OcF2;

• We were hoping for a synthetic solution. – Lucian Nov 27 '18 at 16:40