Incircle of $\triangle ABC$ touch $AC$ at $D$. $BD$ intersect incircle at $E$. Points $F,G$ on incircle are such points, that $FE \parallel BC,GE \parallel AB$. $I_1,I_2$ are incenters of $DEF,DEG$. Prove that angle bisector of $\angle GDF$ passes though the midpoint of $I_1I_2 $.
Problem
Source: St Petersburg Olympiad 2016, Grade 9, P6
Tags: geometry, incircle, incenter, Petersburg, Grade 9
AlastorMoody
13.03.2019 08:58
Saint Petersburg Olympiad 2016 Grade 9 P6 wrote: Incircle of $\triangle ABC$ touch $AC$ at $D$. $BD$ intersect incircle at $E$. Points $F,G$ on incircle are such points, that $FE \parallel BC,GE \parallel AB$. $I_1,I_2$ are incenters of $DEF,DEG$. Prove that angle bisector of $\angle GDF$ passes though the midpoint of $I_1I_2 $. We'll first provide the proof for a similar type of problem: Saint Petersburg Olympiad 2016 Grade 11 P5 wrote: Incircle of $\triangle ABC$ touch $AC$ at $D$. $BD$ intersect incircle at $E$. Points $F,G$ on incircle are such points, that $FE \parallel BC,GE \parallel AB$. $I_1,I_2$ are incenters of $DEF,DEG$. Prove that $I_1I_2 \perp $ bisector of $\angle ABC$
Let $(I)$ touch $AB,AC$ at $C_1,A_1$. Let $M_1,M_2,M_3$ be midpoints of arc $GD,DF,ED$ not containing $E,E,G$, respectively. Let $I_1,I_2$ be the incenters of $\Delta DEF$,$\Delta DEG$
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[/asy][/asy]
Since, $IA_1 \perp BC$ $\implies$ $IA_1 \perp EF$, therefore, $A_1,C_1$ are midpoints of arc $EF,EG$ not containing $G,F$, respectively. Since, $EC_1DA_1$ is harmonic quadrilateral, hence,
$$-1=(E,D;C_1,A_1) \implies \frac{C_1E}{DC_1}=\frac{A_1E}{A_1D} \implies \frac{I_2C_1}{DC_1}=\frac{A_1I_1}{A_1D} \implies \frac{DI_2}{C_1I_2}=\frac{DI_1}{A_1I_1} \implies I_1I_2 ||A_1C_1$$And since, $BI \perp A_1C_1$ $\implies$ $BI \perp I_1I_2$
Back to Main Problem: Define the following extra points together with the other points defined in the above problem, Let $(I)$ touch $AB,AC$ at $C_1,A_1$. Let $M_1,M_2,M_3$ be midpoints of arc $GD,DF,ED$ not containing $E,E,G$, respectively. Let $I_1,I_2$ be the incenters of $\Delta DEF$,$\Delta DEG$, And Let $I_3,I_4$ be the incenters of $\Delta FEG$, $\Delta FGD$, then from Sharygin CR 2018 P13, we have that, $I_1I_2I_3I_4$ form a rectangle, Lemma: Let $AD$ be the $A-$symmedian chord in $\Delta ABC$, Let $\omega_B$ and $\omega_C$ centered at $B$ and $C$ with radius $DB$ and $DC$ and Let $\omega_B , $ $\omega_C $ $\cap $ $\odot (ABC)$ $=$ $E$ $,F$, respectively, then, the angle bisectors of $\Delta EFD$ and bisector of $\angle EAF$ are concurrent
Let $M_{EF},M_{EF}'$ be the midpoints of arc $EF$ not containing $A$ and containing $A$, and Let $I$ be the incenter of $\Delta EFD$
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Proof: We have, by power of point $IE \cdot IC=IF \cdot IB$ and,
$$-1=(A,D;B,C) \implies \frac{AB}{DB}=\frac{AC}{CD} \implies \frac{AB}{AC}=\frac{\sin C}{\sin B}=\frac{DB}{DC}=\frac{IB}{IC}$$Hence, $IB \sin B=IC \sin C$, now, $$AI=\frac{IE \cdot \sin B}{\sin \angle IAE} , AI=\frac{IF \cdot \sin C}{\sin \angle IAF}$$Hence, now using the fact that $\frac{IF}{IE}=\frac{IC}{IB}$ we have, $\angle IAE=\angle IAF$ and hence, angle bisector of $\angle EAF$ passes through incenter of $\Delta EFD$
Hence, $I_3,I_4$ lie on the bisector of $\angle GDF$ and since, $I_3I_4$ being the diagonal of $I_1I_2I_3I_4$ bisects $I_1I_2 \implies$ bisector of $\angle GDF$ bisects $I_1I_2$
nguyenhaan2209
29.09.2019 11:34
Both: Let D,E,F be tangent point, AD-(I)=G,GK,GL//AB,AC. GDL=180-DAC-ADB=ADC-DAC=2ADE so I1 on DE, I2 on DF. By EI1/FI2=EG/FG=ED/FD so I1I2 perp AI. By KDH=KDF+FDH=FDG+EDA=EDF=HDL so DH bisect I1I2 q.e.d
rafaello
22.04.2021 21:55
Lol, this problem is harder than St Peterburg 2016, grade 11, p5 Let $(I)$ touch the $AB$, $AC$ at $R$, $S$, respectively. Anyway, the crux of the problem is to notice that $I_1I_2\parallel RS$. Now we claim this. However this is quite obvious as $RESD$ is a harmonic bundle, which essentially means that $$\frac{RE}{RD}=\frac{SE}{SD}\implies \frac{RI_2}{RD}=\frac{SI_1}{SD}\implies I_1I_2\parallel RS.$$ Now let $K$ be the midpoint of $RS$, we claim that $K$ lies on the angle bisector of $\angle GDF$. This is obvious, $$\angle KDS=\angle RED=\angle ARE=\angle REG=\angle RDG$$and $$\angle KDR=\angle SDE=\angle ASE=\angle FES=\angle FDS,$$where we used the fact that $RESD$ is a harmonic bundle. A homothety finishes.