Leicich wrote: In an acute triangle $ABC$, let $D$ be a point in $BC$ such that $AD$ is the angle bisector of $\angle{BAC}$. Let $E \neq B$ be the point of intersection of the circumcircle of triangle $ABD$ with the line perpendicular to $AD$ drawn through $B$. Let $O$ be the circumcenter of triangle $ABD$. Prove that $E$, $O$, and $A$ are collinear. Let $O$ be the circumcenter of triangle $ABC$?

Dear Mathlinkers, considering (O) and (ABD), the Reim's theorem applied two times gives the answer... Sincerely Jean-Louis

yunxiu wrote: Leicich wrote: In an acute triangle $ABC$, let $D$ be a point in $BC$ such that $AD$ is the angle bisector of $\angle{BAC}$. Let $E \neq B$ be the point of intersection of the circumcircle of triangle $ABD$ with the line perpendicular to $AD$ drawn through $B$. Let $O$ be the circumcenter of triangle $ABD$. Prove that $E$, $O$, and $A$ are collinear. Let $O$ be the circumcenter of triangle $ABC$? And once again, it had to happen. $O$ is indeed the circumcenter of $ABC$, not $ABD$ (thank you for pointing it out).

[asy][asy] /* Argentina Cono Sur TST 2014 Problem 5, free script by liberator, 22 September 2014 */ unitsize(3cm); pointpen=black; pen dsp = rgb(0.4,0.6,0.8); pen rcp = rgb(0.9,0,0); pen gcp = rgb(0,0.6,0); /* Initialize objects */ pair O = origin; pair A = dir(110); pair B = dir(200); pair C = dir(340); pair Ap = dir(-A); pair D1 = bisectorpoint(B,A,C); pair D = IntersectionPoint(Line(A,D1, 2014,2014), B--C); pair E = IntersectionPoint(Line(A,O,-0.01, 2014), circumcircle(A,B,D)); /* Draw objects */ draw(A--B--C--cycle, dsp+linewidth(1)); draw(A--Ap, dsp); draw(D--E, dsp); draw(B--E, dsp); draw(A--D, dsp); draw(unitcircle, gcp); draw(circumcircle(A,B,D), rcp); /* Place dots on and label each point */ Drawing("A", A, dir(A)); Drawing("B", B, dir(B)); Drawing("C", C, dir(C)); Drawing("A'", Ap, dir(Ap)); Drawing("D", D, dir(-80)); Drawing("E", E, dir(0)); Drawing("O", O, dir(60)); [/asy][/asy] Let $A' \equiv AE \cap (O)$. By Reim's theorem on $(O)$ and $(ABD)$, $DE \parallel A'C$. $\angle DAC + \angle ADE = \angle DAB + \angle ABE = 90^{\circ}$, so $DE \perp CA$. It follows $A'$ is antipodal to $A$ w.r.t $(O)$, so $E,O,A$ are collinear.

We are going to show that $\angle{(AC;AO)}=\angle{(AC;AE)}$ Calling $\angle{DAE}=\alpha$ and $\angle{CAE}=\beta$ we deduce that $\angle{ABC}=\angle{EBA}+\angle{EBC}=90-\beta-\alpha+\alpha=90-\beta$ So by central angle $\angle{AOC}=180-2\beta$ and $\angle{CAO}=\angle{ACO}=\beta$ So $\angle{CAE}=\angle{CAO}=\beta$ and we are done..

Dear Mathlinkers, see also http://www.artofproblemsolving.com/community/c6t48f6h1294020_hong_kong_2017_tst1_p1 Sincerely Jean-Louis

Let $P$ be the intersection of $AE$ and the circumcircle of $\triangle ABC$. Also, let $\angle DAC = \varphi$ and $\angle ACB = \gamma$. We will prove $\angle BPA = \pi/2$. Since $A,C,P,B$ are on a circle $\angle BPA=\gamma$. Also, $\angle BEA = \angle BDA = \varphi + \gamma$. This means $\angle PBE = \pi - (\pi - \gamma - \varphi) - \gamma = \varphi$. Also, since $BE \perp AD$, we have $\angle EBA = \pi/2 - \varphi$. So $\angle PBA = \angle PBE + \angle EBA = \varphi + \pi/2 - \varphi = \pi/2$, as desired.

To prove that $A,O$ and $E$ are collinear, we should simply show that: $\angle BAE = \angle BAO$ (sorry for my bad English) $\angle BAE =\angle BAD+ \angle DAE= \angle BAC/2$ $+\pi/2 - \angle AEB=$ $\angle BAC/2$ $+\pi/2 - \angle ADB$ $=\angle BAC/2$ $+\pi/2 - \pi +\angle ABC +\angle BAC/2$ $= \angle BAC + \angle ABC - \pi/2 = \pi/2 - \angle CAB= \angle BAO$

Sorry, where can I understand how to do geometric draw with Latex?

Solutions should be done in cases $\angle BDE\le90^\circ$, $\angle BDE>90^\circ$ because the configuration of points on the drawing is different and angle $DBE$ can't be computed in the second case as $\angle ABD-\angle ABE$

Let G be on BC such that AG is perpendicular to BC and F be the intersection of BE and AD. Therefore, there's a circle passing through A,G,F and B. Now, let x be the angle of FAG, so GBF = x too. Let y be angle GAC, DAB = x + y since AD is angle bissector. And since ADEB is cyclic DEF = x+y. Looking at triangle BED, EBD + EDB = DEF then x + EDB = x + y and we conclude that EDB = y. As ADEB is cyclic EDB = EAB then EAB = y. So, we have: GAC = EAB = y as it's known, orthocenter and circumcenter are conjugates and this ends the proof.

Attachments:

Substitute (alpha - theta) by y and theta by x.

Make: $\angle A = 2\alpha$, $\angle DAE = \beta$, so $\angle BAD = \angle DAC = \alpha$ and $\angle EAC = \alpha - \beta$. Too we have that: $\angle ABE = 90º - \angle BAD = 90º - \alpha$, as $ABDE$ is cyclic $\implies \angle EBD = \beta \implies \angle AOC = 2.(90º - \alpha + \beta),$ so $\angle OAC = \frac{180º - 2.(90º - \alpha + \beta)}{2} = \alpha - \beta$. Therefore $\angle EAC = \angle OAC$ and $A, O, E$ are collinear. $Q.E.D.$

This was also JBMO TST Bosnia and Herzegovina 2019. P2

Let $AK\perp BC$ where $K\in BC$ and let $AD\cap BE=M$. So, as $AMKB$ is a cyclic quadrilateral, so $\angle KAD=\angle MBD=\angle DAE\implies \angle BAK=\angle CAE$. So, $AK$ and $AE$ are isogonal. Also, $AK$ and $AO$ are isogonal as $H\in AK$ where $H$ is the orthocenter of $\triangle ABC$. Hence $E\in AO\implies A,O,E$ are collinear.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Le%20theoreme%20de%20Reim%201.pdf p. 3-4. Sincerely Jean-Louis

WLOG $AB>AC$ and $F$ is the intersection of $BE$ and $AD$ $$\angle DAO= \angle BAD - \angle BAO= \angle BAD + \angle C - 90$$$$\angle DAE= \angle FAE= 90- \angle AEF=\angle AEB - 90=\angle ADB - 90= \angle BAD + \angle C - 90$$

Leicich wrote: In an acute triangle $ABC$, let $D$ be a point in $BC$ such that $AD$ is the angle bisector of $\angle{BAC}$. Let $E \neq B$ be the point of intersection of the circumcircle of triangle $ABD$ with the line perpendicular to $AD$ drawn through $B$. Let $O$ be the circumcenter of triangle $ABC$. Prove that $E$, $O$, and $A$ are collinear. (Using liberator's diagram) Let $E \equiv O \cap \odot(ABD)$. We see that $$\angle BAD = \angle DAC = \angle DAO + \angle OAC = \angle BDE + 90 - \angle ABC = 90 - \angle EBA$$as desired. The other case can be handled similarly.