seoneo wrote: Let $O$ be the outer center of an acute triangle $ABC$. Let $D$ be the intersection of the bisector of the angle $A$ and $BC$. Suppose that $\angle ODC = 2 \angle DAO$. The circumcircle of $ABD$ meets the line segment $OA$ and the line $OD$ at $E (\neq A,O)$, and $F(\neq D)$, respectively. Let $X$ be the intersection of the line $DE$ and the line segment $AC$. Let $Y$ be the intersection of the bisector of the angle $BAF$ and the segment $BE$. Prove that $\frac{\overline{AY}}{\overline{BY}}= \frac{\overline{EX}}{\overline{EO}}$. $O$ is excenter of $\triangle ABC$?

Nice Angle Chasing Problem! [asy][asy] import graph; size(20cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -9.49107455859086, xmax = 14.724907387343215, ymin = -5.142542474435421, ymax = 7.914069629664625; pen qqffff = rgb(0,1,1); pen ffxfqq = rgb(1,0.4980392156862745,0); pen ffqqff = rgb(1,0,1); pen xfqqff = rgb(0.4980392156862745,0,1); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen zzttqq = rgb(0.6,0.2,0); draw((-0.96,6.95)--(-4.087857443145002,0.5576764390231403)--(8.127629121328358,1.4487271675348927)--cycle, linewidth(0.8) + ffxfqq); draw((-2.418823854253927,1.0477001837554554)--(-4.087857443145002,0.5576764390231403)--(-0.96,6.95)--cycle, linewidth(0.8) + green); draw((1.6095493062662123,2.230419641551855)--(1.98,1.55)--(3.0112773986928434,4.545953777169747)--cycle, linewidth(0.8) + green); draw(circle((1.98,1.55), 6.14846322262726), linewidth(0.8) + qqffff); draw((-0.96,6.95)--(-4.087857443145002,0.5576764390231403), linewidth(0.8) + ffxfqq); draw((-4.087857443145002,0.5576764390231403)--(8.127629121328358,1.4487271675348927), linewidth(0.8) + ffxfqq); draw((8.127629121328358,1.4487271675348927)--(-0.96,6.95), linewidth(0.8) + ffxfqq); draw((xmin, -3.404524574352636*xmin + 3.68165640862147)--(xmax, -3.404524574352636*xmax + 3.68165640862147), linewidth(0.8) + ffqqff); draw(circle((-1.8326756606980168,3.4155979832456502), 3.640543973750618), linewidth(0.8) + xfqqff); draw((-0.96,6.95)--(1.98,1.55), linewidth(0.8) + blue); draw((xmin, 0.544135430255349*xmin + 0.4726118480944089)--(xmax, 0.544135430255349*xmax + 0.4726118480944089), linewidth(0.8) + qqffff); draw((xmin, 1.651913911213199*xmin-0.42841724825285465)--(xmax, 1.651913911213199*xmax-0.42841724825285465), linewidth(0.8) + red); draw((xmin, 4.045930424727737*xmin + 10.834093207738627)--(xmax, 4.045930424727737*xmax + 10.834093207738627), linewidth(0.8) + ffqqff); draw((-4.087857443145002,0.5576764390231403)--(1.6095493062662123,2.230419641551855), linewidth(0.8) + yqqqyq); draw((-2.418823854253927,1.0477001837554554)--(-4.087857443145002,0.5576764390231403), linewidth(0.8) + green); draw((-4.087857443145002,0.5576764390231403)--(-0.96,6.95), linewidth(0.8) + green); draw((-0.96,6.95)--(-2.418823854253927,1.0477001837554554), linewidth(0.8) + green); draw((1.6095493062662123,2.230419641551855)--(1.98,1.55), linewidth(0.8) + green); draw((1.98,1.55)--(3.0112773986928434,4.545953777169747), linewidth(0.8) + green); draw((3.0112773986928434,4.545953777169747)--(1.6095493062662123,2.230419641551855), linewidth(0.8) + green); draw(circle((-0.0725977946997371,3.932807867330144), 3.14498506187496), linewidth(0.8) + dotted + zzttqq); draw((-0.96,6.95)--(-0.5112771814188349,0.8185677617691739), linewidth(0.8) + blue); dot((1.98,1.55),dotstyle); label("$O$", (2.131401662221423,1.2811911694581941), NE * labelscalefactor); dot((-0.96,6.95),dotstyle); label("$A$", (-0.7667479352561902,7.12230711076611), NE * labelscalefactor); dot((0.813365769272013,0.9151929809122087),dotstyle); label("$D$", (0.9661662570500114,0.653756720519748), NE * labelscalefactor); dot((-4.087857443145002,0.5576764390231403),linewidth(4pt) + dotstyle); label("$B$", (-4.396904389828665,0.3101616651486941), NE * labelscalefactor); dot((8.127629121328358,1.4487271675348927),linewidth(4pt) + dotstyle); label("$C$", (8.241418081645875,1.2662522540072787), NE * labelscalefactor); dot((1.6095493062662123,2.230419641551855),linewidth(4pt) + dotstyle); label("$E$", (1.1304943270100822,2.4165487437277635), NE * labelscalefactor); dot((-1.1702758875522103,-0.1641767254962735),linewidth(4pt) + dotstyle); label("$F$", (-1.2447932296854871,-0.5562954310043983), NE * labelscalefactor); dot((3.0112773986928434,4.545953777169747),linewidth(4pt) + dotstyle); label("$X$", (3.236881405589172,4.552813653208664), NE * labelscalefactor); dot((-2.418823854253927,1.0477001837554554),linewidth(4pt) + dotstyle); label("$Y$", (-2.335334057602321,0.7433902132252402), NE * labelscalefactor); dot((-0.5112771814188349,0.8185677617691739),linewidth(4pt) + dotstyle); label("$H$", (-0.4082139644342175,0.9525350295380557), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let $H$ be the foot of the perpendicular line through $A$ to side $BC$. First, observe that $\angle HAD = \angle DAO$. This is quite obvious since $O$ is the isogonal conjugate of the orthocenter. Then, $\angle HAO = \angle ODC$ which means $AHDO$ is cyclic. Observe that $\angle AFX = \angle ABC$ and $\angle OAX = 90^\circ - \angle ABC$ Therefore, $\angle AXF = 90^\circ$ Since $\angle AOD = \angle DHA = 90^\circ$, $ADOX$ is cyclic. We have to show that $\frac{\overline{AY}}{\overline{BY}}=\frac{\overline{EX}}{\overline{EO}}$ and it's sufficient to show that $\triangle ABY$ and $\triangle XOE$ are similar! By angle chasing, $$\angle XOE = \angle EDA = \angle EBA = \angle YBA$$and $$\angle BAY = \frac{1}{2} \angle BAF = \frac{1}{2} \angle BDF = \frac{1}{2} \angle ODC = \angle DAO = \angle EXO$$Hence, $\triangle ABY \backsim \triangle XOE$ which ends the proof. $\Box$