ME = MF, intersecting circumcircles, altitudes, angle bisector related

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ME = MF, intersecting circumcircles, altitudes, angle bisector related

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Source: 2013 Romania JBMO TST1 P4

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#1 h May 29, 2020, 10:18 PM

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In the acute-angled triangle $ABC$ , with $AB \ne AC$ , $D$ is the foot of the angle bisector of angle $A$ , and $E, F$ are the feet of the altitudes from $B$ and $C$ , respectively. The circumcircles of triangles $DBF$ and $DCE$ intersect for the second time at $M$ . Prove that $ME = MF$ .

Leonard Giugiuc

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#2 h May 30, 2020, 3:49 AM • 2 Y

Y by Mango247, Mango247

We have that $\angle BFC = \angle BEC = 90$ , so $BFEC$ is cyclic.
Thus, by the Radical Axis Theorem, the circumcircles of $BFEC$ , $\triangle BFD$ , and $\triangle DEC$ concur, so $BF$ , $CE$ , and $DM$ concur, so $DM$ passes through $A$ and is the angle bisector of $\angle BAC$ .
Now $M$ is the Miquel point of $\triangle ABC$ with respect to $\triangle DEF$ , so $\triangle AFME$ is cyclic. Since $\angle EAM = \angle FAM$ , we have $ME=MF$ , as both $ME$ and $MF$ are chords subtending the same angles in the circumcircle of $AFME$ .

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#3 h May 30, 2020, 4:01 AM • 3 Y

Y by Mango247, Mango247, Mango247

Notice that $M$ is a Miquel Point, so $AFEM$ is cyclic. Moreover, the radical axes of $(BFEC), (BFMD),$ and $(CEMD)$ concur at $A.$ But the radical axis of $(BFMD)$ and $(CEMD)$ is $MD$ , so $A,M,D$ are collinear, thus forcing $\angle FAM = \angle EAM,$ so $EM=FM.$

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#4 h Feb 11, 2023, 1:30 PM

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By easy angle chasing we gain that $A,F,M,E$ so M is the miquel point of $ABC$ . Let apply sine theorem on $AFM$ and $AEM$ . Let say circumcircle of $EFBC$ radius is $R$ and so $\frac{\sin FAM}{FM}=\frac{\sin EAM}{ME}=\frac{1}{2R}$ since $\sin FAM=\sin EAM$ we get that $ME=MF$ .

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#5 h Jan 10, 2024, 1:17 PM

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Since $\angle BFC = \angle BEC$ , $BFEC$ is cyclic. Hence $\angle AEF = \angle B$ . Since $M$ is a Maquel Point, $AFME$ is cyclic. $\angle FMD + \angle FMA = 180^{\circ} - \angle B + \angle AEF = 180^{\circ}$ . That is, $M$ lies on $AD$ . So $\angle FAM = \angle EAM$ leading to $EM = FM$ .

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