Perpendicular $KL$ from $I$ to $AD$ meets tangent $BC$ of $(I)$ through $D$ at the pole $P$ of $AD$ WRT $(I)$ $\Longrightarrow$ cross ratio $(B,C,D,P)$ is harmonic. $MN$ is clearly D-midline of $\triangle DIP,$ meeting $DP$ at its midpoint $U,$ hence $UP^2=UD^2=UB \cdot UC,$ or $\tfrac{PU}{UB}=\tfrac{UC}{PU}.$ But by Thales theorem $\tfrac{UC}{UP}=\tfrac{CN}{LN}$ and $\tfrac{PU}{UB}=\tfrac{KM}{BM}$ $\Longrightarrow$ $\tfrac{KM}{BM}=\tfrac{CN}{LN}.$

Let $I$ be the incenter of $\triangle ABC$. $AI$ meet $BC$ at $N$. The incircle touches $BC$ at $D$. Parallel through $I$ to $BC$ meet $AD$ at $P$. $PI = \dfrac {(u-a)(b-c)}{a+b+c}$, where $u$ is semiperimeter. Proof: $CD = u-c$ and $NC = \frac {ab}{b+c} \Longrightarrow DN = \dfrac {(u-a)(b-c)}{b+c}$. $\dfrac{AI}{AN} = \dfrac {AC}{AC+CN} = \dfrac {b}{b + \dfrac {ab}{b+c}} = \dfrac {b+c}{a+b+c}$. $\dfrac {PI}{ DN} = \dfrac {AI}{AN} \Rightarrow PI = DN \cdot \dfrac {AI}{AN} = \dfrac {(u-a)(b-c)}{a+b+c}$. $\blacksquare$ Let's go back to the problem. Let $KL$ meet $AD$ at $R$, and $MN$ meet $AD$ at $S$. Let $X$ be the foot of perpendicular from $B$ to $AD$. Let $Y$ be the foot of perpendicular from $C$ to $AD$. $\dfrac {KM}{BM} = \dfrac {RS}{SX} $ and $\dfrac {LN}{CN} = \dfrac {RS}{SY} $. So $\dfrac {KM\cdot LN}{BM \cdot CN} =1 \Longleftrightarrow RS^2 = SX \cdot SY $ Let $\angle XBD = \theta$. So $\angle ADI = \angle DCY = \theta$. We have $XD = (u-b)\sin\theta$ and $DY=(u-c)\sin\theta$. $SX=SD-XD = RS - XD$ and $SY = SD + DY = RS + DY$ . $SX\cdot SY = (RS-XD)(RS+DY) = RS^2 + RS(DY-XD) - XD\cdot DY$. We will show that $RS(DY-XD) = XD \cdot DY$. $RS = \dfrac r2 \cdot \cos \theta$ where $ID=r$. So $\Longleftrightarrow \dfrac r2 \cdot \cos \theta (b-c)\sin\theta = (u-b)(u-c)\sin^2\theta$ $\Longleftrightarrow r = \dfrac{2(u-b)(u-c)}{(b-c)}\cdot \tan \theta$ $\Longleftrightarrow u(u-a)r = \dfrac{2u(u-a)(u-b)(u-c)}{2(b-c)}\cdot \tan \theta$ $\Longleftrightarrow u(u-a)r = \dfrac{2u^2r^2}{(b-c)}\cdot \tan \theta$ $\Longleftrightarrow (u-a)(b-c) = 2ur \cdot \tan \theta$ $\Longleftrightarrow \dfrac {(u-a)(b-c)}{2u} = r \cdot \tan \theta$ We know $PI = \dfrac {(u-a)(b-c)}{a+b+c}$ and we also know that $PI = r\tan \theta$. So the last statement is true. So we have $RS(DY-XD) = XD \cdot DY$. $\blacksquare$

Luis González wrote: $UP^2=UD^2=UB \cdot UC,$ I don't undestand this part .Can someone explain this more.

bariz kolay soru = obviously easy question

MMEEvN wrote: Luis González wrote: $UP^2=UD^2=UB \cdot UC,$ I don't undestand this part .Can someone explain this more. from Luis González wrote: cross ratio $(B,C,D,P)$ is harmonic. we have $BD*PC=PB*DC$ so $(UP-UB)(UC+UP)=(UP+UB)(UC-UP)$ by opening the brackets the desired conclusion results.

Let $E,F$ be the touch points of the incircle with $AC$ and $AB$ respectively. Let $J$ be the foot from $I$ to $AD$ and let $G$ be the foot from $T$ to $AD$. Now applying Radical Axis Theorem on $(AFJIE) , (DEF) , (JID)$ we get $EF$ , $JI$ and $BC$ concur. Let the concurrency point be $H$. We know that $(H,D;B,C) = -1$. Let $MN\cap BC = Q$. Now since $DT=TI$ we have $DG=GJ$ and because $GQ\parallel HJ$ we have $HQ=QD$. We have $\frac{CN}{NL} = \frac{CQ}{QH} = \frac{CQ}{QD}$ and $\frac{KM}{BM} = \frac{HQ}{QB} = \frac{QD}{QB}$. So it suffices to show that $QD^2 = QB\cdot QC$. Let $BQ=x , BD = y , HQ = x + y , CD = z$. Now using $(H,D;B,C) = -1$ we get $\frac{2x + y}{y} \cdot \frac{z}{z + 2x + 2y} = 1 \implies z = y + \frac{y^2}{x}$. Finally $QD^2 = QB\cdot QC \iff (x+y)^2 = x(x + y + y + \frac{y^2}{x}) = x^2 + 2xy + y^2$. $\square$

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