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Geometry Difficulty 8.7 Shortlist Prove it Taiwan

Let Ω\Omega be the AA-excircle of triangle ABCABC, and suppose that Ω\Omega is tangent to lines BCBC, CACA, and ABAB at points DD, EE, and FF, respectively. Let MM be the midpoint of segment EFEF. Two more points PP and QQ are on Ω\Omega such that EPEP and FQFQ are both parallel to DMDM. Let BPBP meet CQCQ at point XX. Prove that line AMAM is the angle bisector of ∠XAD\angle XAD.

Remark. The AA-excircle is the excircle that lies inside ∠A\angle A.

Solution

Extend DMDM to meet Ω\Omega again at a point RR. Let ARAR meet EFEF and Ω\Omega at UU and V≠RV \ne R respectively; then since FREVFREV is a harmonic quadrilateral, we know
sin⁡∠BARsin⁡∠RAC=sin⁡∠FAUsin⁡∠UAE=FU‾UE‾=FR‾⋅FV‾RE‾⋅VE‾=(FR‾RE‾)2=(FM‾ME‾⋅DE‾FD‾)2 \frac{\sin \angle BAR}{\sin \angle RAC} = \frac{\sin \angle FAU}{\sin \angle UAE} = \frac{\overline{FU}}{\overline{UE}} = \frac{\overline{FR} \cdot \overline{FV}}{\overline{RE} \cdot \overline{VE}} = \left(\frac{\overline{FR}}{\overline{RE}}\right)^2 = \left(\frac{\overline{FM}}{\overline{ME}} \cdot \frac{\overline{DE}}{\overline{FD}}\right)^2

Extend PEPE, QFQF to meet FDFD, DEDE at YY, ZZ respectively; then by the same reasoning we have
sin⁡∠CBPsin⁡∠PBA=−sin⁡∠DBPsin⁡∠PBF=−(DP‾PF‾)2=−(DY‾YF‾⋅EF‾DE‾)2sin⁡∠ACQsin⁡∠QCB=−sin⁡∠ECQsin⁡∠QCD=−(EQ‾QD‾)2=−(EZ‾ZD‾⋅FD‾EF‾)2 \begin{align*} \frac{\sin \angle CBP}{\sin \angle PBA} &= -\frac{\sin \angle DBP}{\sin \angle PBF} = -\left(\frac{\overline{DP}}{\overline{PF}}\right)^2 = -\left(\frac{\overline{DY}}{\overline{YF}} \cdot \frac{\overline{EF}}{\overline{DE}}\right)^2 \\ \frac{\sin \angle ACQ}{\sin \angle QCB} &= -\frac{\sin \angle ECQ}{\sin \angle QCD} = -\left(\frac{\overline{EQ}}{\overline{QD}}\right)^2 = -\left(\frac{\overline{EZ}}{\overline{ZD}} \cdot \frac{\overline{FD}}{\overline{EF}}\right)^2 \end{align*}

Multiplying the three equations together, and noting that EFZYEFZY is a parallelogram, we obtain
sin⁡∠BARsin⁡∠RAC⋅sin⁡∠CBPsin⁡∠PBA⋅sin⁡∠ACQsin⁡∠QCB=(FM‾ME‾⋅DE‾FD‾⋅DY‾YF‾⋅EF‾DE‾⋅EZ‾ZD‾⋅FD‾EF‾)2=(FM‾ME‾⋅DY‾YF‾⋅EZ‾ZD‾)2=(1⋅12⋅2)2=1 \begin{align*} &\frac{\sin \angle BAR}{\sin \angle RAC} \cdot \frac{\sin \angle CBP}{\sin \angle PBA} \cdot \frac{\sin \angle ACQ}{\sin \angle QCB} \\ &= \left( \frac{\overline{FM}}{\overline{ME}} \cdot \frac{\overline{DE}}{\overline{FD}} \cdot \frac{\overline{DY}}{\overline{YF}} \cdot \frac{\overline{EF}}{\overline{DE}} \cdot \frac{\overline{EZ}}{\overline{ZD}} \cdot \frac{\overline{FD}}{\overline{EF}} \right)^2 \\ &= \left( \frac{\overline{FM}}{\overline{ME}} \cdot \frac{\overline{DY}}{\overline{YF}} \cdot \frac{\overline{EZ}}{\overline{ZD}} \right)^2 \\ &= \left( 1 \cdot \frac{1}{2} \cdot 2 \right)^2 = 1 \end{align*}

Therefore, by the trigonometric form of Ceva's theorem, ARAR, BPBP, CRCR are concurrent, that is, AA, XX, RR are collinear.

Let JJ be the AA-excenter of △ABC\triangle ABC. Since AA, MM, JJ are collinear, it suffices to show that AJAJ is the internal angle bisector of ∠RAD\angle RAD. From
MA‾⋅MJ‾=ME‾⋅MF‾=MD‾⋅MR‾ \overline{MA} \cdot \overline{MJ} = \overline{ME} \cdot \overline{MF} = \overline{MD} \cdot \overline{MR}
we know that AA, JJ, DD, RR are concyclic. Moreover, JD‾=JR‾\overline{JD} = \overline{JR}, so AJAJ is the internal angle bisector of ∠RAD\angle RAD, and hence the original statement holds. □

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.