Equal lengths in incircle configuration

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Equal lengths in incircle configuration

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Source: China second round 2018 (A) Q2

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#1 h May 5, 2019, 11:44 AM • 3 Y

Y by Pluto1708, Adventure10, Mango247

In triangle $\triangle ABC$ , $AB<AC$ , $M,D,E$ are the midpoints of $BC$ , the arcs $BAC$ and $BC$ of the circumcircle of $\triangle ABC$ respectively. The incircle of $\triangle ABC$ touches $AB$ at $F$ , $AE$ meets $BC$ at $G$ , and the perpendicular to $AB$ at $B$ meets segment $EF$ at $N$ . If $BN=EM$ , prove that $DF$ is perpendicular to $FG$ .

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#2 h May 5, 2019, 1:35 PM • 2 Y

Y by Adventure10, Mango247

Since $DAGM$ is cyclic, with $\angle DMG = \angle DAG = 90^\circ$ , the conclusion of the problem is equivalent to showing $AFGM$ is cyclic.
Let $DH\perp EF$ at $H$ . Then $FHB \sim EHC$ and $EBIF \sim CDEH$ . Thus $\angle FEI = \angle HCE = \angle HBF$ .
For the original problem, from $\angle ABK = 90^\circ$ we get $D\in KH$ . Let $KD \cap AB = J$ , we have $KBJ \cong NLB \implies BJ= BL$ . From $\frac{BJ}{BA} = \frac{BD}{BG}$ we get $BJ \cdot BG = BA\cdot BD$ , so $BL\cdot BG = BA\cdot BF$ .

Edit: Whoops since the diagram corresponding to the above notes has different notations of the points I should probably state what they are first... Sorry.

Problem: Given acute triangle $ABC$ , $AB < AC$ , let $L$ be the midpoint of $BC$ , $M,N$ the midpoints of arcs $\widehat{BAC}, \widehat{BC}$ respectively. Let $DEF$ be the intouch triangle of $ABC$ and $I$ the incenter of $ABC$ , $G:= AN \cap BC$ , $K\in NF$ such that $KB \perp AB$ . Prove that $BK = LN$ implies $MF \perp FG$ .

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#3 h May 5, 2019, 3:26 PM • 2 Y

Y by Adventure10, Mango247

stroller wrote:

Since $DAGM$ is cyclic, with $\angle DMG = \angle DAG = 90^\circ$ , the conclusion of the problem is equivalent to showing $AFGM$ is cyclic.
Let $DH\perp EF$ at $H$ . Then $FHB \sim EHC$ and $EBIF \sim CDEH$ . Thus $\angle FEI = \angle HCE = \angle HBF$ .
For the original problem, from $\angle ABK = 90^\circ$ we get $D\in KH$ . Let $KD \cap AB = J$ , we have $KBJ \cong NLB \implies BJ= BL$ . From $\frac{BJ}{BA} = \frac{BD}{BG}$ we get $BJ \cdot BG = BA\cdot BD$ , so $BL\cdot BG = BA\cdot BF$ .

Could you check your solution again, define point $K$ and rewrite it?

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#4 h May 5, 2019, 9:00 PM • 1 Y

Y by Adventure10

@above: fixed now. Sorry for the inconvenience.

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#5 h May 6, 2019, 8:19 AM • 1 Y

Y by Adventure10

stroller wrote:

@above: fixed now. Sorry for the inconvenience.

I think it's still not written well!

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#6 h May 10, 2019, 1:40 PM • 6 Y

Y by Pluto1708, albind, mijail, GeoKing, Adventure10, Mango247

Let $I$ be the incenter. Note $\angle NBE=\angle ABE-90^{\circ} =\frac{\angle B-\angle C}{2} = \frac{\angle BEC}{2}-\angle C=\angle MEI$ , combine $EB=EI, BN=EM\implies \triangle NBE\cong \triangle MEI$ . Then $\angle EMI=\angle BNE=180^{\circ}-\angle BNF=180^{\circ}-\angle IFE$ (since $IF\parallel BN\perp AB$ ), thus $F,I,M,E$ are cyclic. Now $\angle EFM=\angle EIM=\angle BEN\implies BE\parallel FM$ , by Reim's theorem (combine $A,D,E,B$ cyclic) we get $A,D,M,F$ cyclic. Thus $A,D,M,G,F$ lie on circle with diameter $DG\implies DF\perp FG$ .

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#7 h Jan 25, 2021, 1:56 AM

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$\because$ point $E$ is the midpoint of the arc $BC$ , point $D$ is the midpoint of the arc $CAB$ ,
$\therefore$ $AE$ is the angle bisector of $\angle BAC$ , $AD$ is the exterior angle bisector of $\angle BAC$ .
Meanwhile, we got $AD \perp AE$ ( $AD \perp AG$ ).
$\therefore$ " $DF \perp FG$ " is equivalent to " $D,A,F,G$ are concyclic".

In addition, " $D,A,F,G$ are concyclic" is equivalent to " $AG \sin \angle FAD = AF \sin \angle GAD + AD \sin \angle FAG$ "(Three-Chord Lemma)
Meanwhile, we got $AE \sin \angle BAD = AF \sin \angle EAD + AD \sin \angle BAE$ .(Three-Chord Lemma)
Combine $\angle FAD = \angle BAD = 90^{\circ} + \frac{A}{2}$ , $\angle GAD = \angle EAD = 90^{\circ}$ , $\angle FAG = \angle BAE = \frac{A}{2}$ ,
We got " $AG \sin \angle FAD = AF \sin \angle GAD + AD \sin \angle FAG$ " is equivalent to " $EG \cos \frac{A}{2} = BF$ ".

$\because$ $EG = \frac{EM}{\sin \angle EGM} = \frac{EM}{\sin(\frac{A}{2}+C)}$ , $BF = BN \tan\angle BNF$ , $EM = BN$
$\therefore$ " $EG \cos \frac{A}{2} = BF$ " is equivalent to " $\frac{\cos\frac{A}{2}}{\sin(\frac{A}{2}+C)} = \tan\angle BNF$ "

Let point $I$ be the incenter of $\triangle ABC$ , we got $IF \perp AB$ ,
$\therefore$ $IF \parallel NB \Leftrightarrow \angle BNF = \angle NFI = \angle EFI$ .
$\therefore$ " $EG \cos \frac{A}{2} = BF$ " is equivalent to " $\frac{\cos\frac{A}{2}}{\sin(\frac{A}{2}+C)} = \tan\angle EFI = \frac{\sin\angle EFI}{\cos\angle EFI}$ "

$\frac{\cos\frac{A}{2}}{\sin(\frac{A}{2}+C)} = \frac{\sin\angle EFI}{\cos\angle EFI} \Leftrightarrow \frac{\cos\angle EFI}{\sin(\frac{A}{2}+C)} = \frac{\sin\angle EFI}{\cos\frac{A}{2}}$
Notice that, $\angle FIE = \angle IAF + \angle AFI = \frac{A}{2} + 90^{\circ}$
$\therefore$ $\cos\frac{A}{2} = \sin(\frac{A}{2} + 90^{\circ}) = \sin\angle FIE$
$\therefore$ $\frac{\sin\angle EFI}{\cos\frac{A}{2}} = \frac{\sin\angle EFI}{\sin\angle FIE} = \frac{EI}{FE} = \frac{EB}{FE} = \frac{\sin\angle EFB}{\sin\angle EBF} = \frac{\sin\angle EFB}{\sin(\frac{A}{2}+B)}$ (The Law of Sine)
In the meantime, we got $(\frac{A}{2} + B) + (\frac{A}{2} + C) = 180^{\circ} \Rightarrow \sin(\frac{A}{2} + B) = \sin(\frac{A}{2} + C)$
Therefore,
$\frac{\cos\angle EFI}{\sin(\frac{A}{2}+C)} = \frac{\sin\angle EFI}{\cos\frac{A}{2}} = \frac{\sin\angle EFB}{\sin(\frac{A}{2}+B)} \Leftrightarrow \cos\angle EFI = \sin\angle EFB$
Finally, we got $\angle EFI = \angle FNB = 90^{\circ} - \angle BFN = 90^{\circ} - \angle EFB \Rightarrow \cos\angle EFI = \sin\angle EFB$
Q.E.D.

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