Nice parallelism

Sashsiam_2

Let $ABC$ be an acute triangle, $AD$ , $BE$ , $CF$ its altitudes, with $D$ in $BC$ , $E$ in $CA$ , and $F$ in $AB$ . Let $M$ be the midpoint of $BC$ . Circumcircle of $AEF$ meets $AM$ at $A$ and at $X$ . $AM$ meets $CF$ at $Y$ . $AD$ meets $BX$ at $Z$ . Prove that $YZ$ is parallel to $BC$ .

**Posted:** Fri Jul 25, 2008 7:48 pm

pohoatza

Denote by $T$ the intersection of $BX$ and $CY$ . Since the segments $EF$ and $BC$ are antiparallel and since the $A$ -symmedian of triangle $AEF$ is the locus of the midpoints of the segments intercepted by the antiparallels to $EF$ with $AB$ and $AC$ , it follows that $AM$ is the $A$ -symmedian of $AEF$ . In this case, the quadrilateral $AEXF$ is harmonic, and thus
$(YZ \cap BC, B, M, C)=Y(Z,B,X,Y)=(Z,B,X, T)$
$=H(Z,B,X, T)=H(A,E,X,F)=(A,E,X,F)=-1$ . We conclude that $YZ \cap BC = \infty$ , i.e. $YZ \| BC$ .

**Posted:** Sat Jul 26, 2008 12:32 am

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Cosmin Pohoata, Bucharest, Romania

plane geometry · Riemann Hypothesis Offline Joined: 06 May 2008 Posts: 468

Denote H , O is the orthocentre and circumcenter of △ABC , respectively
P is midpoint of AH , Q is the center of nine-point circle
MP is the diameter of nine-point circle
MF^2=MX*MA=MB^2 => △BMX∽△AMB => ∠BXM=∠ABC => YXHZ are concyclic =>
YZ⊥AD => YZ∥BC

**Posted:** Sat Jul 26, 2008 1:30 am