Since B and B' are symmetric, their angle bisectors meet on the angle bisector of A Similarly for C and C'. As bisector of B and C meet at I, bisectors of B' and C' meet at I.

Simply isogonal conjugates. Trivial.

yeah it was simple....but i m not confident of the checkers....they usually want some rigorous work.....

@pjmaths agreed

was a synthetic solution necessary?

it was easier to solve by isogonal conjugates without synthetic geometry

idk whats isogonal conjugates.. but it was easy even with normal stuff

can u tell me how u do that

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -16.6775, xmax = 7.782727272727274, ymin = -5.3886363636363574, ymax = 9.884090909090892; /* image dimensions */ pen ffwwqq = rgb(1,0.4,0); /* draw figures */ draw((-4.45,-2.18)--(-1.014447329193773,6.288775901045421), linewidth(0.4)); draw((-1.014447329193773,6.288775901045421)--(2.47,-2.16), linewidth(0.4)); draw((2.47,-2.16)--(-4.45,-2.18), linewidth(0.4)); draw((-3.040398952081422,1.294723437117269)--(2.47,-2.16), linewidth(0.8) + linetype("4 4") + gray); draw((1.0403375621669626,1.306517473285617)--(-4.45,-2.18), linewidth(0.8) + linetype("4 4") + gray); draw((-1.014447329193773,6.288775901045421)--(-0.99,-2.17), linewidth(0.8) + linetype("4 4") + ffwwqq); draw((-3.040398952081422,1.294723437117269)--(1.0403375621669626,1.306517473285617), linewidth(0.4)); draw((-3.040398952081422,1.294723437117269)--(-1.001918473200225,1.9537917272780487), linewidth(0.4)); draw((-1.001918473200225,1.9537917272780487)--(1.0403375621669626,1.306517473285617), linewidth(0.4)); draw((-1.001918473200225,1.9537917272780487)--(-4.45,-2.18), linewidth(0.4)); draw((2.47,-2.16)--(-1.001918473200225,1.9537917272780487), linewidth(0.4)); /* dots and labels */ dot((-4.45,-2.18),dotstyle); label("$C'$", (-4.9843181818181845,-2.320454545454544), NE * labelscalefactor); dot((2.47,-2.16),dotstyle); label("$C$", (2.5327272727272722,-1.996590909090908), NE * labelscalefactor); dot((-1.014447329193773,6.288775901045421),dotstyle); label("$A$", (-0.9445454545454562,6.457954545454534), NE * labelscalefactor); dot((-0.99,-2.17),linewidth(4pt) + dotstyle); label("$D$", (-0.9275,-2.030681818181817), NE * labelscalefactor); dot((-3.040398952081422,1.294723437117269),dotstyle); label("$B$", (-3.296818181818184,1.5829545454545412), NE * labelscalefactor); dot((1.0403375621669626,1.306517473285617),linewidth(4pt) + dotstyle); label("$B'$", (1.10090909090909,1.446590909090905), NE * labelscalefactor); dot((-0.9963097965317255,0.013189599976970714),linewidth(4pt) + dotstyle); label("$E$", (-0.9275,0.1511363636363616), NE * labelscalefactor); dot((-1.001918473200225,1.9537917272780487),linewidth(4pt) + dotstyle); label("$F$", (-0.9275,2.094318181818177), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] It is easy to note that $B'\in AC$ and $C'\in AB$ making triangles $ABB'$ and $ACC'$ isosceles. Also, symmetry says $BC, B'C'$ meet on $AD$ at $E$, where $AD$ is the angle bisector of $\angle A$. Also, $EC=EC'$ and $AC=AC'\implies$ angle bisectors of $\angle AC'E$ and $\angle ACE$ concur at a point $F$ on $AD$ by angle bisector theorem. Thus, $F$ is the incenter of $\triangle ABC$ and $\triangle AB'C'$.

Jupiter_is_BIG wrote: [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -16.6775, xmax = 7.782727272727274, ymin = -5.3886363636363574, ymax = 9.884090909090892; /* image dimensions */ pen ffwwqq = rgb(1,0.4,0); /* draw figures */ draw((-4.45,-2.18)--(-1.014447329193773,6.288775901045421), linewidth(0.4)); draw((-1.014447329193773,6.288775901045421)--(2.47,-2.16), linewidth(0.4)); draw((2.47,-2.16)--(-4.45,-2.18), linewidth(0.4)); draw((-3.040398952081422,1.294723437117269)--(2.47,-2.16), linewidth(0.8) + linetype("4 4") + gray); draw((1.0403375621669626,1.306517473285617)--(-4.45,-2.18), linewidth(0.8) + linetype("4 4") + gray); draw((-1.014447329193773,6.288775901045421)--(-0.99,-2.17), linewidth(0.8) + linetype("4 4") + ffwwqq); draw((-3.040398952081422,1.294723437117269)--(1.0403375621669626,1.306517473285617), linewidth(0.4)); draw((-3.040398952081422,1.294723437117269)--(-1.001918473200225,1.9537917272780487), linewidth(0.4)); draw((-1.001918473200225,1.9537917272780487)--(1.0403375621669626,1.306517473285617), linewidth(0.4)); draw((-1.001918473200225,1.9537917272780487)--(-4.45,-2.18), linewidth(0.4)); draw((2.47,-2.16)--(-1.001918473200225,1.9537917272780487), linewidth(0.4)); /* dots and labels */ dot((-4.45,-2.18),dotstyle); label("$C'$", (-4.9843181818181845,-2.320454545454544), NE * labelscalefactor); dot((2.47,-2.16),dotstyle); label("$C$", (2.5327272727272722,-1.996590909090908), NE * labelscalefactor); dot((-1.014447329193773,6.288775901045421),dotstyle); label("$A$", (-0.9445454545454562,6.457954545454534), NE * labelscalefactor); dot((-0.99,-2.17),linewidth(4pt) + dotstyle); label("$D$", (-0.9275,-2.030681818181817), NE * labelscalefactor); dot((-3.040398952081422,1.294723437117269),dotstyle); label("$B$", (-3.296818181818184,1.5829545454545412), NE * labelscalefactor); dot((1.0403375621669626,1.306517473285617),linewidth(4pt) + dotstyle); label("$B'$", (1.10090909090909,1.446590909090905), NE * labelscalefactor); dot((-0.9963097965317255,0.013189599976970714),linewidth(4pt) + dotstyle); label("$E$", (-0.9275,0.1511363636363616), NE * labelscalefactor); dot((-1.001918473200225,1.9537917272780487),linewidth(4pt) + dotstyle); label("$F$", (-0.9275,2.094318181818177), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] It is easy to note that $B'\in AC$ and $C'\in AB$ making triangles $ABB'$ and $ACC'$ isosceles. Also, symmetry says $BC, B'C'$ meet on $AD$ at $E$, where $AD$ is the angle bisector of $\angle A$. Also, $EC=EC'$ and $AC=AC'\implies$ angle bisectors of $\angle AC'E$ and $\angle ACE$ concur at a point $F$ on $AD$ by angle bisector theorem. Thus, $F$ is the incenter of $\triangle ABC$ and $\triangle AB'C'$. Very well done.. But I think you have to show that $BB'CC'$ is an isoceles trapezium.. I know it's clearly visible, but again, it's also visible that both the triangles will share the same incenter.. Just makes your proof a bit better.. although I never say that this proof is incomplete..

I think we can solve this problem my angle chasing only. So here goes my solution to the problem: WLOG let AC > AB. Let I be the incentre of triangle ABC. Simple angle chasing gives ∠IBB' = ∠ICB = C/2 , ∠ICC' = ∠ABI = B/2 . From which we conclude that points B, C lies on the circumcircle of IBC. Therefore, pentagon BIB'CC' is cyclic which gives ∠IB'A = ∠IBC = ∠ABI = ∠C'CI = ∠C'B'I , ∠IC'B' = ∠ICB' = ∠ICA = ∠BCI = ∠BC'I . This shows that I is also the incentre of triangle AC'B'. Hence proved.