incentres and other concyclic points

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incentres and other concyclic points

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Kyiv mathematical festival

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Source: Kyiv mathematical festival 2014

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#1 h May 9, 2014, 9:06 PM • 1 Y

Y by Adventure10

Let $AD, BE$ be the altitudes and $CF$ be the angle bissector of acute non-isosceles triangle $ABC$ and $AE+BD=AB.$ Denote by $I_A, I_B, I_C$ the incentres of triangles $AEF,$ $BDF,$ $CDE$ respectively. Prove that points $D, E, F, I_A, I_B$ and $I_C$ lie on the same circle.

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#2 h May 9, 2014, 9:48 PM • 1 Y

Y by Adventure10

Perpendicularities $AD \perp BC$ and $BE \perp AC$ are unnecessary; we only need points $D,E$ on $BC,AC,$ such that $AEDB$ is cyclic and $AE+BD=BC.$

Using the converse of 9th ibmo - brazil 1994/q2 (i), it follows that the circle with center $F$ tangent to $CB,CA$ also touches $DE$ $\Longrightarrow$ $F$ becomes C-excenter of $\triangle CDE$ $\Longrightarrow$ $D,E$ are on circle with diameter $\overline{FI_C}.$ Angle chase gives $\widehat{DI_BF}=90^{\circ}+\tfrac{1}{2}\widehat{DBA}=90^{\circ}+\tfrac{1}{2}(180^{\circ}-\widehat{DEA})=180^{\circ}-\widehat{DEF}$ $\Longrightarrow$ $I_B \in \odot(DEF).$ Likewise $I_A \in \odot(DEF),$ hence $D,E,F,I_A,I_B,I_C$ are concyclic.

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