[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(250); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.000000000000000, xmax = 13.50000000000001, ymin = -8.000000000000000, ymax = 4.500000000000005; /* image dimensions */ /* draw figures */ draw((2.226043290043293,3.876138528138530)--(0.000000000000000,-5.000000000000000)); draw((0.000000000000000,-5.000000000000000)--(13.00000000000000,-5.000000000000000)); draw((2.226043290043293,3.876138528138530)--(13.00000000000000,-5.000000000000000)); draw(circle((5.279802840577815,-5.401625718240774), 2.032210276045033)); draw((2.226043290043293,3.876138528138530)--(5.915168779972126,-7.331959557015978)); draw((0.000000000000000,-5.000000000000000)--(4.644436901183503,-3.471291879465571)); draw((5.915168779972126,-7.331959557015978)--(13.00000000000000,-5.000000000000000)); draw((2.226043290043293,3.876138528138530)--(3.287674484143265,-5.000000000000000)); draw((7.271931197012364,-5.000000000000000)--(2.226043290043293,3.876138528138530)); draw((4.644436901183503,-3.471291879465571)--(3.287674484143265,-5.000000000000000)); draw((4.644436901183503,-3.471291879465571)--(7.271931197012364,-5.000000000000000)); draw((7.271931197012364,-5.000000000000000)--(5.915168779972126,-7.331959557015978)); draw((3.287674484143265,-5.000000000000000)--(5.915168779972126,-7.331959557015978)); /* dots and labels */ dot((2.226043290043293,3.876138528138530),dotstyle); label("$A$", (2.289378435517966,3.972368066042483), NE * labelscalefactor); dot((0.000000000000000,-5.000000000000000),dotstyle); label("$B$", (-0.4811843350448048,-5.531306553911202), NE * labelscalefactor); dot((13.00000000000000,-5.000000000000000),dotstyle); label("$C$", (12.95282351756770,-5.499090707741868), NE * labelscalefactor); dot((5.147608451903122,-5.000000000000000),dotstyle); label("$D$", (5.204912513842743,-4.903097553609179), NE * labelscalefactor); dot((4.644436901183503,-3.471291879465571),dotstyle); label("$M$", (4.705566898218057,-3.372844860565789), NE * labelscalefactor); dot((5.915168779972126,-7.331959557015978),dotstyle); label("$N$", (5.865337360314101,-7.770307862679952), NE * labelscalefactor); dot((3.287674484143265,-5.000000000000000),dotstyle); label("$X$", (2.772616128057984,-5.531306553911202), NE * labelscalefactor); dot((7.271931197012364,-5.000000000000000),dotstyle); label("$Y$", (7.331158361018823,-4.903097553609179), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Without loss of generality we can let $X$ be closer to $B$ than $Y$. Since $\triangle ABM \sim \triangle ACN$, $\triangle BMD \sim \triangle CND$, and $AD$ bisects $\angle BAC$, we have \[\frac{AM}{AN} = \frac{AB}{AC} = \frac{BD}{CD} = \frac{MD}{ND}.\] Therefore $(AMDN)$ is harmonic. Because $\angle MXN = \angle MYN = 90^\circ$, it immediately follows that $XM$ bisects $\angle AXD$ and $YM$ bisects $\angle AYD$. Thus $M$ is the incenter of $\triangle AXY$, so $AM$ bisects $\angle XAY$, from which it follows that $\angle BAX = \angle CAY$.

robinpark wrote: Since $\triangle ABM \sim \triangle ACN$, $\triangle BMD \sim \triangle CND$, and $AD$ bisects $\angle BAC$, we have \[\frac{AM}{AN} = \frac{AB}{AC} = \frac{BD}{CD} = \frac{MD}{ND}.\] Therefore $(AMDN)$ is harmonic. Alternatively let $B_1$ be the reflection of $B$ over $M$ (which is on $\overline{AC}$), then project $B$, $B_1$, $M$ from $C$ onto line $AD$.

Note that $BM$ and $CN$ are tangent to the circle, so \[BM^2=BX\cdot BY\qquad\text{and}\qquad CN^2=CY\cdot CX.\]Furthermore, since $\angle BAM=\angle CAM$, we have $\triangle BAM\sim\triangle CAN$. Thus \[\dfrac{BX}{XC}\cdot\dfrac{BY}{YC}=\dfrac{BX\cdot BY}{XC\cdot YC}=\left(\dfrac{MB}{MC}\right)^2=\left(\dfrac{AB}{AC}\right)^2,\]and so by Steiner $AX$ and $AY$ are isogonal as desired. $\blacksquare$

robinpark wrote: Let $AD$ be a bisector of triangle $ABC$. Points $M$ and $N$ are projections of $B$ and $C$ respectively to $AD$. The circle with diameter $MN$ intersects $BC$ at points $X$ and $Y$. Prove that $\angle BAX = \angle CAY$. Let $T$ lie on $BC$ such that $\overline{AT}$ is an external bisector of angle $BAC$. Notice that $$(MN, DA)=(BC, DT)=-1.$$Note if $L$ is the midpoint of $\overline{MN}$ then $$DX \cdot DY=DM\cdot DN=DL \cdot DA$$so $A, X, Y, L$ are concyclic. Meanwhile, $LX=LY$ so by $\overline{AL}$ bisects angle $XAY$, so we are done.

Dear Mathlinkers, see also http://www.artofproblemsolving.com/community/c6t48f6h1280041_interesting_symmedian_problem Sincerely Jean-Louis

This can be easily thoughtlessly barybashed, and as I'm short in time, I'm putting a sketch: Lemma: Let $P_1 = (x_1, y_1, z_1)$ and $P_2 = (x_2, y_2, z_2)$ be the normalized barycentric coordinates of two points $P_1$ and $P_2$ w.r.t triangle $\Delta ABC$, and let $\omega$ be the circle with diameter $P_1P_2$. Then $$ \text{Pow}_{\omega}{A} = \frac{Q - [b^2(z_1+z_2) + c^2(y_1+y_2)]}{2}$$, where $ Q = \sum_{\sigma} a^2 (y_1z_2+z_1y_2)$. (Similarly we get cyclic formulas for $ \text{Pow}_{\omega}{B}$ and $\text{Pow}_{\omega}{C}$)

Let $P$ be the perpendicular from $A$ to $BC$. Ofcourse $A,C,N, P$ anad $A,B, M, P$ are cyclic, so after intersecting the cevian $AI$ with $\omega_{APC}$ and $\omega_{APB}$, We get $M$ and $N$ to be $(b-c: b: c)$ and $(c-b: b: c)$ in some order. So, $$ \text{AX, AY are isogonal w.r.t} \angle BAC \Leftrightarrow \frac{BX*BY}{CX*CY} = \frac{b^2}{c^2} \Leftrightarrow \frac{Pow_{\omega}B}{Pow_{\omega}C} = \frac{b^2}{c^2}$$, where $\omega$ is the circle with diameter $MN$. Now, by the lemma, after normalization and some calculation, we get $Pow_{\omega}{B} = -\frac{b^2[a^2+(b-c)^2]}{4bc}$, which immediately concludes the proof. ( For some reasons the barycentric coordinates of either $M$ or $N$ is same as of the point $P$ in USA December TST for IMO 2015, Problem 1)

Suppose that the perpendicular from $A$ to $AD$ meets $BC$ at $G$. We know that $(G,D;B,C)$ is harmonic. Projecting from the point at infinity onto $AD$ yields $(A,D;M,N)$ harmonic. Projecting through $Y$ onto the circle with diameter $MN$ yields $(YA\cap MN, X;M,N)$ harmonic. As $MN$ is a diameter, this implies that $YA \cap MN = Z$ is the reflection of $X$ in $MN$. Thus $\angle{MAZ} = \angle{MAX} \implies \angle{BAX} = \angle{CAY}$.

Does anyone has an invertive solution using $\sqrt{bc}$

It is very well known that for two points $X$ and $Y$ lying on side $BC$ of $\triangle ABC$, $AX$ and $AY$ are isogonal iff $\frac{BX}{XC}\cdot \frac{BY}{YC}=\left(\frac{AB}{AC}\right)^2$ Since $\angle BAM=\angle CAN$ and $\angle AMB=\angle ANC=90^{\circ}$ it follows that $\triangle ABM\sim \triangle ACN$ and $BM$ and $CN$ are tangent to $(MN)$ Now, by PoP, we see $$\frac{BX\cdot BY}{CX\cdot CY}=\left(\frac{BM}{CN}\right)^2=\left(\frac{AB}{AC}\right)^2$$and hence $AX$ and $AY$ are isogonal and $\angle BAX=\angle CAY$.$\blacksquare$

Firstly, observe that $\tfrac{BM}{CN}=\tfrac{[ABD]}{[ACD]}=\tfrac{BD}{CD}=\tfrac{AB}{AC}.$ Moreover, from PoP, we have that $BM^2=BX\cdot BY$ and similarly $CN^2=CY\cdot CX.$ Hence, $$\left(\frac{AB}{AC}\right)^2=\left(\frac{BM}{CN}\right)^2=\frac{BX\cdot BY}{CX\cdot CY},$$Which reduces to Isogonal Ratios lemma, proving the problem. $\blacksquare$

$X$ is the midpoint of $MN$. $(M ,N$;$ D,A)=-1$ implies $XD\cdot XA=XM^2$. It is well known that inverse of intersection point of diogonals is miquel point. Thus $A$ is miquel point of $MXNY$ which implies the line perpendicular to $AD, NY$ and $MX$ intersect at $T$. $-1=(A,D;M,N)=(AT\cap BC,D;X,Y)$ as desired.

Cute one indeed! I think it's similar to wu2481632's solution.

Let $AE$ be the exterior bisector of $\angle BAC$, with $E\in BC$.Then $(A,D;M,N)\overset{\infty_{BM}}{=}(E,D;B,C)=-1$ so $AE$ is the polar of $D$ wrt $\odot(MN)$.By Brocard, $\{F\}=MX\cap YN$ is on $AE$, hence we have $(X,Y;D,E)\overset{F}{=}(M,N;D,A)=-1$.Thus $\angle XAD=\angle YAD\Rightarrow \angle BAX=\angle CAY$. $\square$

Notice $BM$ and $CN$ are each tangent to $MXNY$. Thus, $$BM^2 = Pow_{(MXNY)}(B) = BX \cdot BY$$and $$CN^2 = Pow_{(MXNY)}(C) = CX \cdot CY.$$It's easy see $ABM \sim ACN$ by AA similarity. Thus, $$\frac{BX}{XC} \cdot \frac{BY}{YC} = \frac{BX \cdot BY}{CX \cdot CY} = \frac{BM^2}{CN^2} = \left(\frac{AB}{AC} \right)^2.$$The result follows via a well-known lemma. $\blacksquare$ Remarks: The well-known lemma is from Chapter 4 of EGMO. It can be proven with the LoS and Phantom Points. Also, I first tried to separately use inversion and apply the Iran Lemma .

Notice that $BM^2=BX\cdot BY,$ $BN^2=CY\cdot CX$ by PoP. Since $\triangle BMD\sim\triangle CND,$ $$\frac{AB^2}{AC^2}=\frac{BD^2}{CD^2}=\frac{BM^2}{CN^2}=\frac{XB}{XC}\cdot\frac{YB}{YC}$$and the result follows by the Isogonal Ratios lemma. $\square$

Too funny to be true View the problem from the "point of view" of lines $\overline{BM}$ and $\overline{CN}$. It becomes the following: Restated problem wrote: Let points $B,C$ lie on two different parallel lines $\ell_1,\ell_2$, and let $\omega$ be a circle tangent to $\ell_1$ at $M$ and $\ell_2$ at $N$. Suppose $\overline{BC}$ intersects $\omega$ at $X,Y$ with $B,X,Y,C$ collinear in that order. Let $A$ lie on $\overline{MN}$ such that $\overline{AMN}$ bisects $\angle BAC$. Prove that it also bisects $\angle XAY$. Let $B',C',X',Y'$ be the reflections of $B,C,X,Y$ over $\overline{MN}$, so $A,B,C'$ and $A,B',C$ are collinear. It suffices to show that $\overline{XY'},\overline{X'Y},\overline{BC'}$ concur. Let $\overline{BC}$ and $\overline{B'C'}$ intersect at $D$, which lies on $\overline{MN}$. Then $$(D,B;X,Y)\stackrel{M}{=}(N,M;X,Y)=(N,M;X',Y')\stackrel{N}{=}(C',D;X',Y')=(D,C';Y',X'),$$whence we are done by Prism Lemma.

WLOG assume $AB < AC$. Take $\sqrt{bc}$ inversion. Let $P'$ denote the image of $P$. Note that $D'$ is the midpoint of arc $BC$ not containing $A$, and the image of $(MN)$ is $(M'N')$. Let $A_1$ be the antipode of $A$ in $(ABC)$. Since $\angle ABN' = \angle ACM' = 90^\circ$, we have that $M'=A_1C \cap AD$ and $N'=A_1B \cap AD$. Now $\angle D'A_1N'= \angle D'A_1B = \angle D'AB = \angle D'AC = 180^\circ - D'A_1C = \angle D'A_1M'$, and $\angle A_1D'M' = 90^\circ = \angle A_1D'N'$, so $\Delta A_1D'M' \cong \Delta A_1D'N' \implies M'D'=N'D'$. So $D'$ is the center of $(M'N')$. It follows that $X'Y' // BC$, so $\angle BAX' = \angle CAY'$. Inverting back gives the desired result.

Let $D \equiv AM \cap BC,$ $J$ is midpoint of $MN$. It's easy to see that $\triangle ABM \sim \triangle ACN$. Then $\dfrac{DM}{DN} = \dfrac{BM}{CN} = \dfrac{AM}{AN}$ or $(MN, DA) = - 1$. Combine with $\angle{MXN} = \angle{MYN} = 90^{\circ},$ we have $XM, YM$ are bisectors of $\angle{AXD}, \angle{AYD}$. Hence $\angle{XJY} = 360^{\circ} - 2\angle{XMY} = 2(\angle{MXY} + \angle{MYX}) = \angle{AXD} + \angle{AYD} = 180^{\circ} - \angle{XAY}$ or $A, X, J, Y$ lie on a circle. But $JX = JY$ then $AJ$ is bisector of $\angle{XAY}$ or $AX, AY$ are isogonal in $\angle{BAC}$

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