This problem was proposed by Burii.

Surprisingly simple with barycentrics. Apparently it is not necessary to even calculate the length of $AM$ with Stewart’s Theorem because the coordinates of $F$ with the reference triangle $ABC$ turn out as $(??? : b(2t-a) : c(2t-a))$ where $t$ denotes the length of $AM$. So clearly $F$ lies on the $A$-angle bisector. The calculations take fewer than 10 minutes to do. I’d love to see a synthetic solution though. Best, V.A

Ez length bash and trigonometry

pggp wrote: Let $M$ be the midpoint of the side $BC$ of an acute triangle $ABC$. The incircles of triangles $ABM$ and $ACM$ are tangent to sides $AB, AC$ at $D, E$ respectively. Let $F$ be the point such that the quadrilateral $DMEF$ is a parallelogram. Prove that $F$ lies on the bisector of $\angle BAC$. We will prove the more general result: Claim. Let $D$ and $E$ be points on $AB$ and $AC$ such that $AD-AE=\tfrac{1}{2}(AB-AC)$. Let $F$ be the point such that $DMEF$ is a parallelogram. Then $AF$ bisects angle $BAC$. Proof. WLOG, assume $AB<AC$. Animate $D$ linearly on $AB$. Notice that $D \mapsto E$ and hence $D \mapsto F$ are linear maps. When $D$ coincides with the midpoint of $AB$, we see that $F=A$. So we only need to check the case $D=B$, to finish the proof. Indeed, $E$ lies on $AC$ such that $CE=\tfrac{1}{2}(AC-AB)$. Reflect $C$ in $E$ to get the point $G$. Then $ME \parallel BG$ and $AG=AB$. Thus, $F$ lies on $BG$ with $EF \parallel BC$, hence $F$ is the midpoint of $BG$, proving that $AF$ bisects angle $BAC$, as desired. $\blacksquare$ The problem follows since $AD-AE=\tfrac{1}{2}(AB+AM-BM)-\tfrac{1}{2}(AC+AM-CM)=\tfrac{1}{2}(AB-AC)$, as desired. $\blacksquare$

Expanding on the idea of #5, here is what seems to be yet a simpler proof for the anantmudgal09 wrote: Claim. Let $D$ and $E$ be points on $AB$ and $AC$ such that $AD-AE=\tfrac{1}{2}(AB-AC)$. Let $F$ be the point such that $DMEF$ is a parallelogram. Then $AF$ bisects angle $BAC$. Indeed, consider the points $D', E',F', M'$ obtained by scaling with a factor of $2$ with center $A$. Note that $ABM'C$ and $D'M'E'F'$ are parallelograms. Moreover $BD'=CE'$. Now clearly $F'$ is obtained by shifting $A$ by the sum of vectors $CE'$ and $BD'$ so that $F'$ clearly lies on the angular bisector.

Proof using barycentrics: let $t=AM$. Obviously, $$M=\left(0:\frac{1}{2}:\frac{1}{2} \right)$$We have $BD= \frac{\frac{a}{2} + c + t}{2}$ and $AD = \frac{c+t - \frac{a}{2}}{2}$. Thus, $$D= \left( \frac{\frac{a}{2} + c + t}{2} : \frac{c+t - \frac{a}{2}}{2} : 0 \right) = \left( \frac{a+2c-2t}{4c} : \frac{2c+2t-a}{4c} : 0 \right)$$Similarly, we obtain $$E = \left(\frac{a+2b-2t}{4b}:0:\frac{2b+2t-a}{4b} \right)$$We calculate $F$: $$F=\left( \frac{a+2b-2t}{4b} + \frac{a+2c-2t}{4c} : \frac{2c+2t-a}{4c} - \frac{1}{2} : \frac{2b+2t-a}{4b} - \frac{1}{2} \right) = \left( \frac{a+2b-2t}{4b} + \frac{a+2c-2t}{4c} : \frac{2t-a}{4c} : \frac{2t-a}{4b} \right) $$$F$ lies on the bisector of $\angle BAC$ because $$\frac{\frac{2t-a}{4c}}{\frac{2t-a}{4b}}=\frac{b}{c} \ \square$$

What about these two ideas? 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anantmudgal09 wrote: pggp wrote: Let $M$ be the midpoint of the side $BC$ of an acute triangle $ABC$. The incircles of triangles $ABM$ and $ACM$ are tangent to sides $AB, AC$ at $D, E$ respectively. Let $F$ be the point such that the quadrilateral $DMEF$ is a parallelogram. Prove that $F$ lies on the bisector of $\angle BAC$. We will prove the more general result: Claim. Let $D$ and $E$ be points on $AB$ and $AC$ such that $AD-AE=\tfrac{1}{2}(AB-AC)$. Let $F$ be the point such that $DMEF$ is a parallelogram. Then $AF$ bisects angle $BAC$. Proof. WLOG, assume $AB<AC$. Animate $D$ linearly on $AB$. Notice that $D \mapsto E$ and hence $D \mapsto F$ are linear maps. When $D$ coincides with the midpoint of $AB$, we see that $F=A$. So we only need to check the case $D=B$, to finish the proof. Indeed, $E$ lies on $AC$ such that $CE=\tfrac{1}{2}(AC-AB)$. Reflect $C$ in $E$ to get the point $G$. Then $ME \parallel BG$ and $AG=AB$. Thus, $F$ lies on $BG$ with $EF \parallel BC$, hence $F$ is the midpoint of $BG$, proving that $AF$ bisects angle $BAC$, as desired. $\blacksquare$ The problem follows since $AD-AE=\tfrac{1}{2}(AB+AM-BM)-\tfrac{1}{2}(AC+AM-CM)=\tfrac{1}{2}(AB-AC)$, as desired. $\blacksquare$ can you explain more about your solution ( using linear map) ???

timon92 wrote: What about these two ideas? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.42cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.42, xmax = 6., ymin = -1.06, ymax = 5.24; /* image dimensions */ /* draw figures */ draw((-0.88,4.54)--(1.8,-0.38), linewidth(1.2)); draw(circle((-0.29337118523817657,0.8714249290319653), 1.239710772398044), linewidth(1.2)); draw(circle((2.409817104810321,0.6400959928637379), 1.0234868209384973), linewidth(1.2)); draw((-1.78,-0.36)--(1.8,-0.38), linewidth(1.2)); draw((0.01055865050020059,-0.27000156046408175)--(0.009441349499799393,-0.4699984395359185), linewidth(1.2)); draw((1.8,-0.38)--(5.38,-0.4), linewidth(1.2)); draw((3.5905586505002,-0.29000156046408176)--(3.5894413494997983,-0.48999843953591854), linewidth(1.2)); draw((5.38,-0.4)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); 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dot((3.0438483981234756,1.4435445548354677),linewidth(4.pt) + dotstyle); label("$E$", (3.12,1.6), NE * labelscalefactor); dot((-1.2453704047551948,2.5507611296661667),dotstyle); label("$B'$", (-1.6,2.68), NE * labelscalefactor); dot((0.7076967962469514,3.2870891096709354),dotstyle); label("$C'$", (0.78,3.48), NE * labelscalefactor); dot((-0.2688368042541217,2.9189251196685513),linewidth(4.pt) + dotstyle); label("$F$", (-0.3,2.5), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.42cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.46, xmax = 5.96, ymin = -1.04, ymax = 5.26; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw(arc((-0.88,4.54),0.6,-100.40771131249006,-69.34300109626494)--(-0.88,4.54)--cycle, linewidth(1.2) + qqwuqq); draw(arc((-0.88,4.54),0.6,-69.34300109626493,-38.278290880039776)--(-0.88,4.54)--cycle, linewidth(1.2) + red); draw(arc((3.0438483981234756,1.4435445548354677),0.6,110.65699890373509,141.72170911996025)--(3.0438483981234756,1.4435445548354677)--cycle, linewidth(1.2) + red); draw(arc((2.432685202377597,3.064619435166917),0.6,-100.40771131249006,-69.34300109626494)--(2.432685202377597,3.064619435166917)--cycle, linewidth(1.2) + qqwuqq); /* draw figures */ draw((-0.88,4.54)--(-1.78,-0.36), linewidth(1.2)); draw((-1.78,-0.36)--(5.38,-0.4), linewidth(1.2)); draw((-0.88,4.54)--(1.8,-0.38), linewidth(1.2)); draw(circle((-0.29337118523817657,0.8714249290319653), 1.239710772398044), linewidth(1.2)); draw(circle((2.409817104810321,0.6400959928637379), 1.0234868209384973), linewidth(1.2)); draw((-1.5126852023775974,1.0953805648330834)--(-0.2688368042541216,2.9189251196685513), linewidth(1.2)); draw((-0.2688368042541216,2.9189251196685513)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); draw((3.0438483981234756,1.4435445548354677)--(1.8,-0.38), linewidth(1.2)); draw((1.8,-0.38)--(-1.5126852023775974,1.0953805648330834), linewidth(1.2)); draw((-0.88,4.54)--(2.432685202377597,3.064619435166917), linewidth(1.2)); draw((2.432685202377597,3.064619435166917)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); draw((-0.2688368042541216,2.9189251196685513)--(-0.88,4.54), linewidth(1.2)); draw((-0.88,4.54)--(2.25,2.07), linewidth(1.2)); draw((2.25,2.07)--(2.432685202377597,3.064619435166917), linewidth(1.2)); draw((2.235761823752463,2.5460329829670103)--(2.4324712569988693,2.509902678901344), linewidth(1.2)); draw((2.2502139453787287,2.6247167562655727)--(2.4469233786251356,2.5885864521999062), linewidth(1.2)); draw((2.25,2.07)--(1.8,-0.38), linewidth(1.2)); draw((2.25,2.07)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); draw((2.6774719171298393,1.860052657552742)--(2.5535755893331076,1.7030504284014183), linewidth(1.2)); draw((2.740272808790368,1.8104941264340495)--(2.6163764809936367,1.6534918972827257), linewidth(1.2)); draw((3.0438483981234756,1.4435445548354677)--(5.38,-0.4), linewidth(1.2)); /* dots and labels */ dot((-0.88,4.54),dotstyle); label("$A$", (-0.8,4.74), NE * labelscalefactor); dot((-1.78,-0.36),dotstyle); label("$B$", (-1.88,-0.78), NE * labelscalefactor); dot((5.38,-0.4),dotstyle); label("$C$", (5.34,-0.8), NE * labelscalefactor); dot((1.8,-0.38),linewidth(4.pt) + dotstyle); label("$M$", (1.72,-0.76), NE * labelscalefactor); dot((-1.5126852023775974,1.0953805648330834),linewidth(4.pt) + dotstyle); label("$D$", (-1.92,1.06), NE * labelscalefactor); dot((3.0438483981234756,1.4435445548354677),linewidth(4.pt) + dotstyle); label("$E$", (3.12,1.6), NE * labelscalefactor); dot((-0.2688368042541216,2.9189251196685513),linewidth(4.pt) + dotstyle); label("$F$", (-0.34,2.5), NE * labelscalefactor); dot((2.432685202377597,3.064619435166917),linewidth(4.pt) + dotstyle); label("$G$", (2.52,3.22), NE * labelscalefactor); dot((2.25,2.07),linewidth(4.pt) + dotstyle); label("$N$", (2.,2.32), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The first one: We can show $F$ lies on $B’C’$ with congruency and phantom point. Thumbs up

Great problem! Let $B'$ be the reflection of $B$ over $D$ and $C'$ be the reflection of $C$ over $E$. Let $F'$ be the intersection of $B'C'$ and the angle bisector of $\angle BAC$. Let the incircle of the triangle $ABM$ be tangent to sides $AB$, $BM$ and $AM$ at $D$, $G$, $J$, respectively. Let the incircle of the triangle $ACM$ be tangent to sides $AC$, $CM$ and $AM$ at $E$, $H$, $K$, respectively. Claim. $AB'=AC'$. $$AB'=AD-BD=AJ-BG=AJ-(MB-MG)=AM-MC=AK-CH=AE-CE=AC'.\quad \square$$ Therefore, we have $F'$ being a midpoint of $B'C'$. Now since $D$, $F'$, $E$, $M$ are the midpoints of quadrilateral $BB'C'C$, by Varignon's theorem, $DMEF'$ is a parallelogram. Hence, we have $F\equiv F'$, since there is unique point such that the quadrilateral $DMEF$ is a parallelogram. $\square$

timon92 wrote: What about these two ideas? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.42cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.42, xmax = 6., ymin = -1.06, ymax = 5.24; /* image dimensions */ /* draw figures */ draw((-0.88,4.54)--(1.8,-0.38), linewidth(1.2)); draw(circle((-0.29337118523817657,0.8714249290319653), 1.239710772398044), linewidth(1.2)); draw(circle((2.409817104810321,0.6400959928637379), 1.0234868209384973), linewidth(1.2)); draw((-1.78,-0.36)--(1.8,-0.38), linewidth(1.2)); draw((0.01055865050020059,-0.27000156046408175)--(0.009441349499799393,-0.4699984395359185), linewidth(1.2)); draw((1.8,-0.38)--(5.38,-0.4), linewidth(1.2)); draw((3.5905586505002,-0.29000156046408176)--(3.5894413494997983,-0.48999843953591854), linewidth(1.2)); draw((5.38,-0.4)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); draw((4.212776926823901,0.3937126317233789)--(4.336673254620632,0.5507148608747027), linewidth(1.2)); draw((4.149976035163372,0.4432711628420715)--(4.273872362960103,0.6002733919933954), linewidth(1.2)); draw((4.087175143502843,0.49282969396076415)--(4.211071471299574,0.6498319231120879), linewidth(1.2)); draw((3.0438483981234756,1.4435445548354677)--(0.7076967962469514,3.2870891096709354), linewidth(1.2)); draw((1.876625324947377,2.237257186558847)--(2.000521652744109,2.394259415710171), linewidth(1.2)); draw((1.8138244332868474,2.28681571767754)--(1.9377207610835794,2.4438179468288634), linewidth(1.2)); draw((1.751023541626318,2.3363742487962322)--(1.8749198694230498,2.4933764779475562), linewidth(1.2)); draw((0.7076967962469514,3.2870891096709354)--(-0.2688368042541217,2.9189251196685513), linewidth(1.2)); draw((0.10714490951527315,3.060674392646202)--(0.16651409346698828,3.243363472771171), linewidth(1.2)); draw((0.10714490951527315,3.060674392646202)--(0.27234589852584185,2.962650756568315), linewidth(1.2)); draw((-0.2688368042541217,2.9189251196685513)--(-1.2453704047551948,2.5507611296661667), linewidth(1.2)); draw((-0.8693886909858001,2.6925104026438174)--(-0.810019507034085,2.8751994827687866), linewidth(1.2)); draw((-0.8693886909858001,2.6925104026438174)--(-0.7041877019752312,2.594486766565931), linewidth(1.2)); draw((-1.2453704047551948,2.5507611296661667)--(-1.5126852023775974,1.0953805648330834), linewidth(1.2)); draw((-1.2734470261300597,1.8443475818660733)--(-1.4701564593764662,1.8804778859317397), linewidth(1.2)); draw((-1.2878991477563262,1.765663808567511)--(-1.484608581002733,1.8017941126331773), linewidth(1.2)); draw((-1.5126852023775974,1.0953805648330834)--(-1.78,-0.36), linewidth(1.2)); draw((-1.540761823752462,0.38896701703298975)--(-1.7374712569988688,0.42509732109865594), linewidth(1.2)); draw((-1.5552139453787286,0.3102832437344274)--(-1.7519233786251354,0.34641354780009365), linewidth(1.2)); draw((-1.2453704047551948,2.5507611296661667)--(-0.88,4.54), linewidth(1.2) + red); draw((-0.88,4.54)--(0.7076967962469514,3.2870891096709354), linewidth(1.2) + red); draw((-0.2688368042541217,2.9189251196685513)--(-0.88,4.54), linewidth(1.2)); draw((-0.2688368042541217,2.9189251196685513)--(-1.5126852023775974,1.0953805648330834), linewidth(1.2)); draw((-1.5126852023775974,1.0953805648330834)--(1.8,-0.38), linewidth(1.2)); draw((1.8,-0.38)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); draw((3.0438483981234756,1.4435445548354677)--(-0.2688368042541217,2.9189251196685513), linewidth(1.2)); /* dots and labels */ dot((-0.88,4.54),dotstyle); label("$A$", (-0.8,4.74), NE * labelscalefactor); dot((-1.78,-0.36),dotstyle); label("$B$", (-1.86,-0.78), NE * labelscalefactor); dot((5.38,-0.4),dotstyle); label("$C$", (5.28,-0.8), NE * labelscalefactor); dot((1.8,-0.38),linewidth(4.pt) + dotstyle); label("$M$", (1.68,-0.74), NE * labelscalefactor); dot((-1.5126852023775974,1.0953805648330834),linewidth(4.pt) + dotstyle); label("$D$", (-1.9,1.08), NE * labelscalefactor); dot((3.0438483981234756,1.4435445548354677),linewidth(4.pt) + dotstyle); label("$E$", (3.12,1.6), NE * labelscalefactor); dot((-1.2453704047551948,2.5507611296661667),dotstyle); label("$B'$", (-1.6,2.68), NE * labelscalefactor); dot((0.7076967962469514,3.2870891096709354),dotstyle); label("$C'$", (0.78,3.48), NE * labelscalefactor); dot((-0.2688368042541217,2.9189251196685513),linewidth(4.pt) + dotstyle); label("$F$", (-0.3,2.5), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.42cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.46, xmax = 5.96, ymin = -1.04, ymax = 5.26; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw(arc((-0.88,4.54),0.6,-100.40771131249006,-69.34300109626494)--(-0.88,4.54)--cycle, linewidth(1.2) + qqwuqq); draw(arc((-0.88,4.54),0.6,-69.34300109626493,-38.278290880039776)--(-0.88,4.54)--cycle, linewidth(1.2) + red); draw(arc((3.0438483981234756,1.4435445548354677),0.6,110.65699890373509,141.72170911996025)--(3.0438483981234756,1.4435445548354677)--cycle, linewidth(1.2) + red); draw(arc((2.432685202377597,3.064619435166917),0.6,-100.40771131249006,-69.34300109626494)--(2.432685202377597,3.064619435166917)--cycle, linewidth(1.2) + qqwuqq); /* draw figures */ draw((-0.88,4.54)--(-1.78,-0.36), linewidth(1.2)); draw((-1.78,-0.36)--(5.38,-0.4), linewidth(1.2)); draw((-0.88,4.54)--(1.8,-0.38), linewidth(1.2)); draw(circle((-0.29337118523817657,0.8714249290319653), 1.239710772398044), linewidth(1.2)); draw(circle((2.409817104810321,0.6400959928637379), 1.0234868209384973), linewidth(1.2)); draw((-1.5126852023775974,1.0953805648330834)--(-0.2688368042541216,2.9189251196685513), linewidth(1.2)); draw((-0.2688368042541216,2.9189251196685513)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); draw((3.0438483981234756,1.4435445548354677)--(1.8,-0.38), linewidth(1.2)); draw((1.8,-0.38)--(-1.5126852023775974,1.0953805648330834), linewidth(1.2)); draw((-0.88,4.54)--(2.432685202377597,3.064619435166917), linewidth(1.2)); draw((2.432685202377597,3.064619435166917)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); draw((-0.2688368042541216,2.9189251196685513)--(-0.88,4.54), linewidth(1.2)); draw((-0.88,4.54)--(2.25,2.07), linewidth(1.2)); draw((2.25,2.07)--(2.432685202377597,3.064619435166917), linewidth(1.2)); draw((2.235761823752463,2.5460329829670103)--(2.4324712569988693,2.509902678901344), linewidth(1.2)); draw((2.2502139453787287,2.6247167562655727)--(2.4469233786251356,2.5885864521999062), linewidth(1.2)); draw((2.25,2.07)--(1.8,-0.38), linewidth(1.2)); draw((2.25,2.07)--(3.0438483981234756,1.4435445548354677), linewidth(1.2)); draw((2.6774719171298393,1.860052657552742)--(2.5535755893331076,1.7030504284014183), linewidth(1.2)); draw((2.740272808790368,1.8104941264340495)--(2.6163764809936367,1.6534918972827257), linewidth(1.2)); draw((3.0438483981234756,1.4435445548354677)--(5.38,-0.4), linewidth(1.2)); /* dots and labels */ dot((-0.88,4.54),dotstyle); label("$A$", (-0.8,4.74), NE * labelscalefactor); dot((-1.78,-0.36),dotstyle); label("$B$", (-1.88,-0.78), NE * labelscalefactor); dot((5.38,-0.4),dotstyle); label("$C$", (5.34,-0.8), NE * labelscalefactor); dot((1.8,-0.38),linewidth(4.pt) + dotstyle); label("$M$", (1.72,-0.76), NE * labelscalefactor); dot((-1.5126852023775974,1.0953805648330834),linewidth(4.pt) + dotstyle); label("$D$", (-1.92,1.06), NE * labelscalefactor); dot((3.0438483981234756,1.4435445548354677),linewidth(4.pt) + dotstyle); label("$E$", (3.12,1.6), NE * labelscalefactor); dot((-0.2688368042541216,2.9189251196685513),linewidth(4.pt) + dotstyle); label("$F$", (-0.34,2.5), NE * labelscalefactor); dot((2.432685202377597,3.064619435166917),linewidth(4.pt) + dotstyle); label("$G$", (2.52,3.22), NE * labelscalefactor); dot((2.25,2.07),linewidth(4.pt) + dotstyle); label("$N$", (2.,2.32), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Do you have any solution for your second idea?

Choose $G$ so that $\vec{AG}=\vec{DM}=\vec{FE}$. Then $AD \parallel GM$ and $AF \parallel GE$. Let $N = AC \cap MG$. Then $N$ is the midpoint of $AC$ because $AB \parallel MN$ and $M$ is the midpoint of $BC$. We have $\angle DAF = \angle MGE$ and $\angle FAE = \angle GEA$. So we have to prove that $\angle MGE = \angle GEA$. This is equivalent to $NG = NE$. We have $NG = MG - MN = AD - \frac{AB}{2} = \frac{AB+AM-BM}{2}-\frac{AB}{2} = \frac{AM}{2} - \frac{BC}{4}$ and $EN = AE - AN = \frac{AC+AM-MC}{2}-\frac{AC}{2} = \frac{AM}{2} - \frac{BC}{4}$. This shows that $NG=NE$ and we are done.

We will prove $dist(F,AB)=dist(F,AC)$ It is easy to se that $dist(F,AB)=dist(M,AB)-dist(D,AB)-dist(E,AB)=dist(M,AB)-dist(E,AB)$ Let $\angle BAM = \alpha$ $\angle CAM =\beta$ $BC=2a$ $AC=2b$ $AB=2c$ $dist(F,AB)=AMsin(\alpha)-AEsin(\alpha + \beta)=AMsin(\alpha)-\frac{AM+2b-a}{2}sin(\alpha + \beta)$ $dist(M,AC)=AMsin(\beta)-ADsin(\alpha + \beta)=AMsin(\beta)-\frac{AM+2c-a}{2}sin(\alpha + \beta)$ We should prove that $AM(sin(\alpha)-sin(\beta))=(b-c)sin(\alpha+\beta)$ equivalent to $\frac{AM}{sin(\alpha+\beta)}=\frac{b-c}{sin(\alpha)-sin(\beta)}$ By $A(\triangle ABM)+A(\triangle AMC)=A(\triangle ABC)$ we have $AM.c.sin(\alpha)+AM.b.sin(\beta)=2bcsin(\alpha+\beta)$ $\rightarrow$ $\frac{AM}{sin(\alpha+\beta)}=\frac{2bc}{csin(\alpha)+bsin(\beta)}$ We should prove that $\frac{b-c}{sin(\alpha)-sin(\beta)}=\frac{2bc}{csin(\alpha)+bsin(\beta)}$ which is true by $\frac{sin(\alpha)}{sin(\beta)}=\frac{b}{c}$ $\blacksquare$