The following generalized is true. rkm0959 wrote: A acute triangle $\triangle ABC$ has circumcenter $O$. The circumcircle of $OAB$, called $O_1$, and the circumcircle of $OAC$, called $O_2$, meets $BC$ again at $D ( \not=B )$ and $E ( \not= C )$ respectively. Let $T$ be the circumcenter of $\triangle ADE$ and let the perpendicular bisector of $BC$ intersect $O_1O_2$ at $K$. Prove that $A,K,T$ collinear. My solution uses some painful and lengthy angle chasing.

Let the angles of $\triangle ABC $ be $A,B,C $. We shall show that both conditions are equivalent to $2B=90^{\circ}+C $. Suppose centres of $O_1,O_2$ and $F $ are collinear. Since the radical axis of $O_1,O_2$ is $AO $ we conclude that the perpendicular bisectors of $AO,BC $ intersect on $AC $ at $F $. Considering $F $ as the intersection of perpendicular bisector of $AO$ with $AC $, we have $\angle OFC=2\angle OAF=2 (90^{\circ}-B) $. Considering $F $ as the intersection of perpendicular bisector of $BC $ with $AC $, we get $\angle OFC=90^{\circ}-C $. Thus, $2 (90^{\circ}-B)=90^{\circ}-C\implies 2B=90^{\circ}+C $, as required. Suppose the circumcentre of $\triangle ADE $ lie on $AC $ then $D$ lies on segment $BC $ while $E$ lies on $BC $ extended so $B>A>C $. Now $\angle ADE = A-C+B $ by simple angke chase while $\angle EAC=B-A $. Hence if the circumcentre lies on $AC $, then $\angle ADE+\angle EAC=90^{\circ}\implies (A-C+B)+(B-A)=90^{\circ}\implies 2B=90^{\circ}+C $, as required. Hence we are done.

babu2001 wrote: Let the angles of $\triangle ABC $ be $A,B,C $. We shall show that both conditions are equivalent to $2B=90^{\circ}+C $. Suppose centres of $O_1,O_2$ and $F $ are collinear. Since the radical axis of $O_1,O_2$ is $AO $ we conclude that the perpendicular bisectors of $AO,BC $ intersect on $AC $ at $F $. Considering $F $ as the intersection of perpendicular bisector of $AO$ with $AC $, we have $\angle OFC=2\angle OAF=2 (90^{\circ}-B) $. Considering $F $ as the intersection of perpendicular bisector of $BC $ with $AC $, we get $\angle OFC=90^{\circ}-C $. Thus, $2 (90^{\circ}-B)=90^{\circ}-C\implies 2B=90^{\circ}+C $, as required. Suppose the circumcentre of $\triangle ADE $ lie on $AC $ then $D$ lies on segment $BC $ while $E$ lies on $BC $ extended so $B>A>C $. Now $\angle ADE = A-C+B $ by simple angke chase while $\angle EAC=B-A $. Hence if the circumcentre lies on $AC $, then $\angle ADE+\angle EAC=90^{\circ}\implies (A-C+B)+(B-A)=90^{\circ}\implies 2B=90^{\circ}+C $, as required. Hence we are done. Exactly same with me

You can actually just prove both implications, but yes, it is easily equivalent to $2B=90+C$. Same as mine.

We prove ThE-dArK-lOrD's generalisation. ThE-dArK-lOrD wrote: A acute triangle $\triangle ABC$ has circumcenter $O$. The circumcircle of $OAB$, called $O_1$, and the circumcircle of $OAC$, called $O_2$, meets $BC$ again at $D ( \not=B )$ and $E ( \not= C )$ respectively. Let $T$ be the circumcenter of $\triangle ADE$ and let the perpendicular bisector of $BC$ intersect $O_1O_2$ at $K$. Prove that $A,K,T$ collinear. Claim. $O_1,O_2$ are the midpoints of arcs $\widehat{AD},\widehat{AE}$. Proof. Since $AO=CO$, $EO$ bisects $\angle DEA$. But $O_1A=O_1D$, so $O_1$ must be the arc midpoint; similarly, $O_2$ is an arc midpoint. $\square$ Now we use complex numbers with $(ADE)$ as the unitcircle. Let $a^2,d^2,e^2$ be the complex numbers of $A,D,E$. Then $o_1=-ad$, $o_2=-ae$ and $o=-(ab+bc+ca)$. Redefine $K=AT\cap O_1O_2$; then by the chord intersection formula, $k=\frac{-a^3(b+c)}{a^2+bc}$. Now $$\frac{k-o}{b^2-c^2}=\frac{bc(a+b)(a+c)}{(b+c)(b-c)\left(a^2+bc\right)}=-\frac{\frac{1}{bc}\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a}+\frac{1}{c}\right)}{\left(\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{bc}\right)}=-\overline{\left(\frac{k-o}{b^2-c^2}\right)},$$so $OK\perp BC$, and the result follows.

ThE-dArK-lOrD wrote: A acute triangle $\triangle ABC$ has circumcenter $O$. The circumcircle of $OAB$, called $O_1$, and the circumcircle of $OAC$, called $O_2$, meets $BC$ again at $D ( \not=B )$ and $E ( \not= C )$ respectively. Let $T$ be the circumcenter of $\triangle ADE$ and let the perpendicular bisector of $BC$ intersect $O_1O_2$ at $K$. Prove that $A,K,T$ collinear. I'll prove ThE-dArK-lOrD's generalization: Claim 1: $O$ is the incenter of $\triangle ADE$. Proof. $AO = BO$, so $O$ is the midpoint of arc $\widehat{AB}$ of $\odot(AOB)$. Thus, $DO$ bisects $\angle ADE$. Similarly $EO$ bisects $\angle ADE$. $\square$ Claim 2: $O_1, O_2$ are the midpoints of arcs $\widehat{AD}$ and $\widehat{AE}$ of $\odot(ADE)$, respectively. Proof. \[\measuredangle AOO_1 = 90^{\circ} + \measuredangle OAB = \measuredangle ACB = \measuredangle AOE\]So, $EOO_1$ are collinear. Along with this, $O_1$ lies on the perpendicular bisector of $AD$, so, $O_1$ is the midpoint of arc $\widehat{AD}$. $\square$ [asy][asy] unitsize(2.5 cm); import geometry; import olympiad; pair A = dir(100), B = dir(200), C = -1/B, O = (0, 0); circle c_1 = circle((point) A,(point) B,(point) O), c_2 = circle((point) A, (point) C,(point) O); pair D = intersectionpoints(line(B, C), c_1)[1], E = intersectionpoints(line(B, C), c_2)[0]; pair O_1 = circumcenter(A, B, O), O_2 = circumcenter(A, C, O); pair K = extension(O_1, O_2, (B+C)/2, O); pair T = circumcenter(A, D, E); draw(A--B--C--cycle); dot(A^^B^^C); dot(O^^O_1^^O_2); draw(c_1^^c_2, linewidth(0.4)+dashed); dot(D^^E); draw(D--A--E--cycle, linewidth(1)); dot(K); draw(circumcircle(A, D, E), dashed+linewidth(0.7)); dot(T); draw(O_1--O_2); draw(A--K--T, linewidth(1.1)); draw(K--(B+C)/2); label("$A$", A, A); label("$B$", B, B); label("$C$", C, C); label("$D$", D, dir(240)); label("$E$", E, dir(-60)); label("$O$", O, dir(0)); label("$O_1$", O_1, dir(150)); label("$O_2$", O_2, dir(60)); label("$K$", K, dir(150)); label("$T$", T, dir(-60)); [/asy][/asy] So the problem now transforms into: Quote: $ABC$ is a triangle with incenter $I$ and circumcenter $O$. $Y, Z$ are the intersection points of $\odot(ABC)$ with $BI$ and $CI$, respectively. Let $K$ be the intersection point of $AO$ with $YZ$. Prove that $KI \perp BC$. From here, it's very easy to complex bash it. I'll provide a synthetic solution. Let $X$ be the intersection of $AI$ and $\odot(ABC)$. Note that $YZ$ is the perpendicular bisector $AI$. So, $KA = KI$. Also, clearly, $OA = OX$. Therefore, we have \[\measuredangle KIA = \measuredangle IAK = \measuredangle XAO = \measuredangle AXO\]Therefore, $IK \parallel OX$. Since, $X$ is the midpoint of arc $\widehat{BC}$, $OX \perp BC$. Thus, $IK \perp BC$. [asy][asy] unitsize(2.5 cm); import geometry; pair A = dir(110), B = dir(210), C = -1/B, O = (0, 0), I = incenter(A, B, C); pair X = circumcenter(I, B, C), Y = circumcenter(I, C, A), Z = circumcenter(I, A, B); pair K = extension(A, O, Y, Z), D = projection(B, C)*I; draw(unitcircle); draw(A--B--C--cycle); draw(X--Y--Z--cycle); dot(A^^B^^C); dot(X^^Y^^Z); dot(O^^I); dot(K); draw(O--A--X--O); draw(K--D, linewidth(1.2)); markangle(I, A, K, n=2, radius=0.7 cm); markangle(K, I, A, n=2, radius=0.7 cm); markangle(O, X, A, n=2, radius=0.7 cm); markrightangle(K, D, B); markrightangle(O, (B+C)/2, C); label("$A$", A, A); label("$B$", B, B); label("$C$", C, C); label("$X$", X, X); label("$Y$", Y, Y); label("$Z$", Z, Z); label("$O$", O, dir(0)); label("$I$", I, -dir(0)); label("$K$", K, dir(-30)); [/asy][/asy]

[asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -20.13839182578881, xmax = 24.678453195216232, ymin = -20.864724143538304, ymax = 16.934536603715053; /* image dimensions */pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw(arc((3.9048088393944984,3.417712349159264),1.196179137571309,-151.68988080175083,-114.63464561747348)--(3.9048088393944984,3.417712349159264)--cycle, linewidth(2) + qqwuqq); draw(arc((3.9048088393944984,3.417712349159264),1.196179137571309,-88.16226320961216,-51.10702802533482)--(3.9048088393944984,3.417712349159264)--cycle, linewidth(2) + qqwuqq); /* draw figures */draw(circle((4.1440446669087585,-4.038470941701886), 7.460020311506091), linewidth(0.8) + zzttff); draw(circle((-3.4620030175639,-0.5505855990750161), 8.367634405361127), linewidth(0.8) + qqwuqq); draw((3.9048088393944984,3.417712349159264)--(-1.662239610071297,-8.722375930381483), linewidth(0.8) + fuqqzz); draw((11.24983850229547,-1.766996306786579)--(-1.662239610071297,-8.722375930381483), linewidth(0.8)); draw((11.24983850229547,-1.766996306786579)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw(circle((6.839384262002882,-0.22005980400870417), 4.673876244547078), linewidth(0.8) + qqwuqq); draw(circle((1.7195675443454639,-1.347653441134675), 5.24251758536787), linewidth(0.8) + linetype("4 4") + blue); draw((-3.4620030175639,-0.5505855990750161)--(6.839384262002882,-0.22005980400870417), linewidth(0.8)); draw((4.1440446669087585,-4.038470941701886)--(3.9048088393944984,3.417712349159264), linewidth(0.8) + fuqqzz); draw((-3.4620030175639,-0.5505855990750161)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw((6.839384262002882,-0.22005980400870417)--(3.9048088393944984,3.417712349159264), linewidth(0.8)); draw((-3.4620030175639,-0.5505855990750161)--(5.704572246887101,-4.754077982147205), linewidth(0.8)); draw((6.839384262002882,-0.22005980400870417)--(2.3715829157948893,-6.549467219835816), linewidth(0.8)); /* dots and labels */dot((4.1440446669087585,-4.038470941701886),dotstyle); label("$O$", (2.987738167256496,-4.3175794071352005), NE * labelscalefactor); dot((3.9048088393944984,3.417712349159264),dotstyle); label("$A$", (3.9845541152325867,4.733509400487703), NE * labelscalefactor); dot((-3.4620030175639,-0.5505855990750161),dotstyle); label("$O_1$", (-5.0665346923903165,-0.21069770147370676), NE * labelscalefactor); dot((-1.662239610071297,-8.722375930381483),linewidth(4pt) + dotstyle); label("$C$", (-2.35519531389535,-9.979493991639396), NE * labelscalefactor); dot((2.16791200013861,-0.369946614549803),linewidth(4pt) + dotstyle); label("$F$", (1.0339789092233584,0.1481560397976859), NE * labelscalefactor); dot((11.24983850229547,-1.766996306786579),linewidth(4pt) + dotstyle); label("$B$", (12.676789181584098,-1.7657305803164083), NE * labelscalefactor); dot((6.839384262002882,-0.22005980400870417),linewidth(4pt) + dotstyle); label("$O_{2}$", (7.493346252108426,-1.0878957356926666), NE * labelscalefactor); dot((2.3715829157948893,-6.549467219835816),linewidth(4pt) + dotstyle); label("$E$", (2.1105401330375364,-7.906116819849127), NE * labelscalefactor); dot((5.704572246887101,-4.754077982147205),linewidth(4pt) + dotstyle); label("$D$", (5.579459631994332,-6.191593389330251), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] CLAIM. $A,O_1,O_2,E,D$ are concyclic. Proof. Notice that by angle chasing we easily have $EA=EB$ and $DA=DC$. Hence $E,O,O_2$ and $D,O,O_1$ are collinear. Hence $\angle O_2O_1D=\frac{1}{2}\angle OO_1A=\angle ACO=\angle CAO=\angle OED=\angle O_2ED$. Meanwhile $\angle O_2O_1D=\angle ACO=90^{\circ}-\angle ABD=\angle O_2AD$ as desired. $\blacksquare$ $\noindent\rule{15.5cm}{0.1pt}$ Now \begin{align*} &\text{The circumcenter of }\triangle AED\hspace{1pt} \text{lies on} AC\\ \iff &\angle O_1AC=\angle O_2AO\\ \iff &90^{\circ}-\angle AEB=90^{\circ}-\angle ABO\\ \iff &2B=C+90^{\circ}\\ \iff &\angle AFO=\angle AOC\\ \iff &\triangle FAO\sim \triangle OAC\\ \iff &FA=FO\\ \iff &F \text{lies on the perpendicular bisector of} AO=O_1O_2 \end{align*}as desired.

We show that both conditions are equivalent to $\angle C + 90^\circ = 2\angle B$. First Restriction. Suppose $F$ lies on $\overline{O_1O_2}$. Then $\overline{O_1F}$ is a perpendicular bisector of $\overline{AO}$, so $AO=OF$. But $$\angle AFO = 180^\circ- (90^\circ-\angle C) = \angle C + 90^\circ$$and $\angle OAC = 90^\circ - B$, so $$\angle C + 90^\circ + 180^\circ - 2\angle B = 180^\circ \iff \angle C + 90^\circ = 2\angle B.$$ Second Restriction. Suppose the circumcenter $O_3$ of $\triangle ADE$ lies on $\overline{AC}$. Then $$\angle O_3AD = \angle A - \angle BAD = \angle A - 180^\circ + \angle B + 2\angle C = \angle C.$$Because $\angle AO_3D = 2\angle AED = 360^\circ - 4\angle B$, $$360^\circ - 4\angle B + 2\angle C = 180^\circ \iff \angle C + 90^\circ + 2\angle B.$$Thus both statements are equivalent to the same statement, and they must be equivalent.