WLOG $AB<AC$. Let $F$ be the reflection of $D$ over $BE$. Note that $F$ is the orthocenter of $ABE$, so it suffices to show that $\angle ABF=\angle C$ by circumcenter angles. But $\angle ADB=180^\circ-\frac{\angle A}{2}-\angle B$, so $\angle EBD=\frac{\angle A}{2}+\angle B-90^\circ$, so $\angle FBD=\angle A+2\angle B-180^\circ=\angle B-\angle C$, as desired.

WLOG, assume that $\triangle ABC$ is acute and $AB>AC$. Let $P=DE\cap AC, Q=BE\cap AD$. Then we have that $\angle DAP=\angle BAD=\angle DEQ$, so $A, P, Q, E$ are concyclic. Hence $\angle AQE=90^\circ\implies \angle APD=90^\circ$. Now $\angle BAE=\angle BDE=\angle CDP=90^\circ-\angle ACB\implies O\in AE$.

Let angles $A,B,C$ be equal to $2\alpha, 2\beta, 2\gamma$ so $\angle OAB=\alpha+\beta-\gamma$ $\angle DAO=\beta-\gamma$ $\angle DAE=\angle DBE=2\beta-\angle ABE=2\beta-(180-90-\alpha)=2\beta-90+\alpha=\beta-\gamma$. The rest is trivial.

Dear Mathlinkers, have a look at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=605458 Sincerely Jean-Louis

Simple angle chasing using cyclic quads.

This doesn't work.

@2above Doesn't Quote: $$\angle BOE=180^\circ-\angle OBE-\angle OEB$$$$180^\circ-\left(\angle C-\frac{\angle A}{2}\right)-\angle ADB$$ assume what we want to prove?

No, because it only uses the fact that $BOE$ is a triangle but doesn't assume that $O$ and $E$ are on the same line as $A$. Triangles $AOE$ and $BOE$ don't have to have their bases on the same line.

Then how do you get $\angle{OEB} = \angle{ADB}$?

See the proposed problem PP19 from here

@Generic_Username: Wait, never mind. You're right. It's $\angle AEB=\angle ADB$ but then my solution wouldn't work.

Let $H$ be the orthocenter of $\triangle ABC$, and $H_b$ the orthocenter of $ABD$. Then $A,H,H_b$ collinear on the perpendicular from $A$ to $BC$. Then $AO$ is the reflection of $AH$ in $AD$, and $AE$ is the reflection of $AH_b$ in $AD$, giving the desired result.

Have never done Hong Kong TSTs before, so it's hard for me to judge, whether this is an easy or average #1. Let's see, how rusty I am with directed angles, it's been ages since I've used them. So here is a simple angle chasing solution. Let $S = AD \cap BE$ and $L=DE \cap CA$. Use directed angles modulo $180^{\circ}$. It is easy to show that $\measuredangle BAO = 90^{\circ}-\measuredangle ACB$. Also, $\measuredangle BAE = \measuredangle BDE = \measuredangle CDE$ in cyclic quadrilateral $ABDE$. But $\measuredangle BAD = \measuredangle DAL$ as $AD$ is an angle bisector and \[ \measuredangle SBA = \measuredangle EBA = \measuredangle EDA = \measuredangle LDA \]in cyclic quadrilateral $ABDE$. So $\triangle ADL \sim \triangle ABS$ and therefore $\measuredangle ALD = \measuredangle ASB = 90^{\circ}$. Thus \[ \measuredangle BAE = \measuredangle CDE = 90^{\circ} - \measuredangle LCD = 90^{\circ}-\measuredangle ACB. \]So $\measuredangle BAO = \measuredangle BAE$ which implies that $A,O,E$ are collinear. Done. $\hfill \square$.

We show that $ \measuredangle BAE = \measuredangle BAO $ $ \measuredangle BAE = \measuredangle BAD +\measuredangle DAE $ $= \measuredangle BAD + \measuredangle DBE $ $ = \measuredangle BAD + \measuredangle ABC - \measuredangle EBA $ $ = \measuredangle BAD +\measuredangle ABC - (90^{\circ} - \measuredangle BAD) $ $ = 2 \measuredangle BAD + \measuredangle ABC -90^{\circ} $ $ =\measuredangle BAC + \measuredangle ABC -90^{\circ} $ $ = \measuredangle BCA -90^{\circ} = \measuredangle BAO $ $\implies$ $ A E O $ collinear.

Dear Mathlinkers, 1. the tangent to (ABD) at B 2. A' the second point of intersection of AE with (O) 3. by applying the Reim's theorem two times we are done... Sincerely Jean-Louis

Let $F $ be the intersection of $AE$ with the circumcircle of $\Delta ABC $. Observe that $DE\perp AC $. We have $\angle EDC = \angle EAB = \angle BCF $. So, $DE || FC $. This gives $\angle ACF = 90^{\circ} $ and hence the result follows.

Argentina Cono Sur TST 2014 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. I'd like to give some warning. Maybe Mister YanYau didn't translate the problem carefully, but we need assumption weaker than in Cono Sur very much: $\angle ACB\le 90^\circ$. It's easy to observe that this equality holds in reverse we have $O$ on the other side of line $AB$ than point $E$, which contradicts collinearity $A,E,O$, because foot of $O$ on segment $AB$ is a center of this segment.

Extend $AD$ to met $(ABC)$ at $M$ By angle chasing we get that $DE$ is perpendicular to $AC$ and by Reim's theorem we have $FC$ is parallel to $DE$ then $\angle AFC$=90.

We can mark the intersection of CA with Circumcircle of ABD as Q and then proceed to prove BDA'=BQC which isn't too hard to do with angle chasing. Explanation: AA' is the diameter of ABC; The problem is equivalent to proving BEO+BEA'=180*.

Slight overkill but we can use areals. Let $P$ be the foot of the perpendicular from $B$ to $AD$ then $P=(k,b,c)$ for some $k$. Using EFFT we see: $$P=(b-c,b,c) \, , \, D=(0,b,c)$$Then solving for circle $\odot ABD$ we get $u=v=0 \, , \, w=\frac{a^2 b}{b+c}$. Letting $E=(b-c,k,c)$ we can solve for $k$ and see: $$k=\frac{b^2 S_B}{c S_C} \implies E=(c S_C (b-c),b^2 S_B,c^2 S_C)$$And as $O=(a^2 S_A,b^2 S_B,c^2 S_C)$ it is then obvious $A,O,E$ are colinear.

What's EFFT?

WolfusA wrote: What's EFFT? Evan’s Favorite Forgotten Trick