IZHO 2020 P4 wrote: In a scalene triangle $ABC$ $I$ is the incentr and $CN$ is the bisector of angle $C$. The line $CN$ meets the circumcircle of $ABC$ again at $M$. The line $l$ is parallel to $AB$ and touches the incircle of $ABC$. The point $R$ on $l$ is such. That $CI \bot IR$. The circumcircle of $MNR$ meets the line $IR$ again at S. Prove that $AS=BS$. Let $MS\cap AB=K$. So, it suffices to show that $SK\perp AB$. Hence, it suffices to show that $INKS$ is a cyclic quadrilateral. So, $\angle RSM=\angle INB=\angle RNM\implies \angle INR=\angle INA$. This is what we are supposed to show in order to prove the conclusion. Now we can safely elimate the points $M,K,S$ and $\odot(MNR)$ from the diagram. So now restate the problem as follows. Restated Problem wrote: $CAB$ is a triangle with $C$ as its apex. Let the line $\ell$ be tangent to $\odot(I)$ and $\|$ to $AB$. Let $R\in\ell$ such that $\angle CIR=90^\circ$. Then show that $\angle ANC=\angle CNR$ For this just note that if $\ell\cap CI=K$ and the tangency point made by $\ell,AB$ with $\odot(I)$ be $X,D$ respectively then $\triangle KXI\cong\triangle NDI\implies \angle KRI=\angle NRI$. So $$\angle CNA=90^\circ-\angle DIN=90^\circ-\angle XIK=90^\circ-\angle XRI=90^\circ-\angle IRN=\angle CNR$$and this is what we were supposed to prove. $\blacksquare$

RnstTrjyn wrote: In a scalene triangle $ABC$ $I$ is the incentr and $CN$ is the bisector of angle $C$. The line $CN$ meets the circumcircle of $ABC$ again at $M$. The line $l$ is parallel to $AB$ and touches the incircle of $ABC$. The point $R$ on $l$ is such. That $CI \bot IR$. The circumcircle of $MNR$ meets the line $IR$ again at S. Prpve that $AS=BS$. $S'\in IR, MS' \perp AB $. We know that $MS'$ is a perpendicular bisector of $AB $. So, it suffices to show that $MNS'R $ is a cyclic. Let $CN \cap l=T, MS' \cap AB=K , IR \cap AB=L$. So, $S'INK$ is a cyclic and $LNRT$ is a romb. Since $\angle INR=\angle ANI=\angle IS'M $. $\implies $ $MNS'R$ is a cyclic quadrilateral. $\blacksquare $

Let $D$ and $D'$ be the tangency point of the incircle with side $BC$ and $l$ respectively and $K=CI\cap l$. Lemma 1:$IN=IK$ It is easy to show that $D,I,D'$ are collinear, $DD' \bot BC$ and $DD'\bot l$. Which implies $\triangle IDN \cong \triangle ID'K$ so, $IN=IK$. Lemma 2: $\angle KID'=\angle KRI$ $\angle KID'+ \angle D'IR=\angle D'RI + \angle D'IR=90$ $\Rightarrow \angle KID'=\angle KRI$ Lemma 3: $ID' \parallel SM$ From, Lemma 2 we get $IN=IK$ and $RI\bot NK$ so, $RK=RN$ and $\angle KRI=\angle IRD$. Now, $$\angle AID'=\angle KRI=\angle IRD=\angle SRD=\angle SMN=\angle SMA$$So, $ID' \parallel SM$. So, $SM$ is perpendicular to $BC$ and $M$ lies on the perpendicular bisector of $BC$. Which implies $AS=BS$.

Let $CM\cap \ell=L$ and $SM\cap AB=K$. Since $AM=BM$, we need to prove that $SM\perp AB \Leftrightarrow \angle SKA=90^\circ \Leftrightarrow \angle NKM=90^\circ$ $\ell$ is parallel to $AB$ and tangent to the incircle of $ABC$ so we get $LI=IN$ then $LR=RN$. $\angle NMK=\angle NMS=\angle NRS=\angle NRI=\angle IRL=90^\circ - \angle ILR=90^\circ - \angle NLR=90^\circ - \angle LNA=90^\circ - \angle MNK$ so $\angle NMK+\angle MNK=90^\circ$ then $\angle NKM=90^\circ$ so we are done.

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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Nice one $\color{black}\rule{25cm}{1pt}$ First let's add some points. Let $S'$ be the intersection of the perpendicular bisector of $AB$ with $RI$, let $N'$ be the intersection of $l$ with $CI$ and let $M'$ be the midpoint of $AB$. We start off with this killer lemma. $\color{red}\rule{25cm}{0.5pt}$ Lemma: $RN$ is tangent to the incircle of $ABC$. Proof (Sketch): Throw the configuration onto the complex plane, with the incircle as the unit circle and let $d=1$, and add the rest of the touch points. Then easily get $r$ since $R\in l$ and $CI \perp RI$. Then calculate $n$ and calculate the touch point (which we call $E$) of the other tangent of $R$ onto the incircle, to get that $RN$ contains $E$. Thus $RN$ is tangent to the incircle of $ABC$. $\color{red}\rule{25cm}{0.5pt}$ Now back to the problem. From our lemma we have that $\angle NRN' = 2 \angle N'RI = 2(90-\angle A- \frac{1}{2}\angle C)$. But this implies that $\angle RNA = \angle NRN'$. Now we have that $\angle RNM = \angle ANR + \angle ANM = \angle B + \frac{1}{2} \angle C$ But notice how we have that $S'M'NI$ is a cyclic quadrilateral, since we have that $\angle S'IN = S'M'N = 90$. This implies that $\angle RS'M = \angle M'NI = \angle B + \frac{1}{2} \angle C$. Thus we have that $RS'NM$ is a cyclic quadrilateral, this implies that $S' \equiv S$. Thus we have that $SA=SB$.

WLOG let $CA > CB.$ First, observe that the reflection of $N$ across $I$ must lie on $\ell$ by homothety. This implies that \[\angle A + \frac{1}{2}\angle B = \angle (\ell, RI) =\angle IRN = \angle SMN,\]so $MS$ must pass through the midpoint of the arc $ACB,$ meaning that $MS$ perpendicularly bisects $AB.$ $\blacksquare$

Let incircle of $ABC$ touches $l$ and $AB$ at $T$ and $D$, respectively. Consider $P = IR \cap AB$. Then easy to see the triangles $TIR$ and $PID$ are congruent which follows $PI = IR$. Therefore segment $NI$ is median and altitude of triangle $PNR$ which implies it's angle bisector, too. So $$\angle INR = \angle INP = \angle TIR = \angle ISM$$where the last one is true b/c of $MNSR$ is cyclic. Then $SM \perp AB$ which gives us required result.

Suppose we cut the perpendicular $IR$ from $M$ to $AB$ at $K$. It is easy to prove that $MNKR$ is cyclic.

Let $L$ be the midpoint of $NR$. Then $IL\parallel AB$ and after that just angle chasing