My solution: Let $ X=AB \cap (P) $ . Since $ \angle CXA+\angle XAC=\frac{\angle CPB+\angle BOC}{2}=90^{\circ} $ , so we get $ \angle ACX=90^{\circ} $ and $ \angle SBF=\angle SCX=\angle CSE $ , hence $ \triangle SBF \sim \triangle CSE $ . Since $ SE=SF $ , so we get $ \frac{SB^2}{SC^2}=\frac{SF}{CE} \cdot \frac{BF}{SE}=\frac{BF}{CE} $ . ... $ (1) $ Since $ SD $ is $ S - $ symmedian of $ \triangle SBC $ , so we get $ \frac{BD}{DC}=\frac{SB^2}{SC^2} $ . ... $ (2) $ From $ (1) $ and $ (2) $ we get $ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=\frac{CE}{FB} \cdot \frac{BD}{DC}=\frac{SC^2}{SB^2} \cdot \frac{SB^2}{SC^2}=1 $ , so from the converse of Ceva theorem we get $ AD, BE, CF $ are concurrent . Q.E.D

andreiromania wrote: Let $\bigtriangleup ABC$ be an acute triangle of circumcentre $O$.Let the tangents to the circumcircle of $\bigtriangleup ABC$ in points $B$ and $C$ meet at point $P$.The circle of centre $P$ and radius $PB=PC$ meets the internal angle bisector of $\angle BAC$ inside $\bigtriangleup ABC$ at point $S$,and $OS \cap BC=D$.The projections of $S$ on $AC$ and $AB$ respectively are $E$ and $F$.Prove that $AD$,$BE$ and $CF$ are concurrent. If there exists any purely synthetic proof I would love to read it. Let's Q be the projection of S on BC, then we have E, F, Q, D concyclic. I have seen this problem several times in AoPS but I can't search any links :-|

qua96 wrote: Let's Q be the projection of S on BC, then we have E, F, Q, D concyclic. My solution: Let $ M $ be the midpoint of $ EF $ . Let $ Y=DE \cap AS, Z=DF \cap AS $ . From the original problem we get $ AD, BE, CF $ are concurrent . Since $ Y(E,F;M,A)=-1 $ So from $ EM=MF $ we get $ BY \parallel EF $ , ie. $ Y $ is the projection of $ B $ on $ AS $ hence we get $ B, F, S, Y $ are concyclic . Similarly, we can prove $ CZ \parallel EF $ and $ C, E, S, Z $ are concyclic . Since $ \angle MEY=90^{\circ}-\angle EYS=90^{\circ}-\angle SYF=90^{\circ}-\angle SBF=\angle FSB $ $ \angle ZFM=90^{\circ}-\angle SZF=90^{\circ}-\angle EZS=90^{\circ}-\angle ECS=\angle CSE $ , so we get $ \angle MED=\angle MZD $ , ie. $ D, E, M, Z $ are concyclic hence $ \angle EDF=90^{\circ} $ . ... $ ( \& ) $ Since $ \angle EQF=\angle EQS+\angle SQF=\angle ECS+\angle SBF=90^{\circ} $ , so combine with $ ( \& ) $ we get $ D, E, F, Q $ are concyclic . Q.E.D

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TelvCohl wrote: My solution: Let $ X=AB \cap (P) $ . Since $ \angle CXA+\angle XAC=\frac{\angle CPB+\angle BOC}{2}=90^{\circ} $ , so we get $ \angle ACX=90^{\circ} $ and $ \angle SBF=\angle SCX=\angle CSE $ , hence $ \triangle SBF \sim \triangle CSE $ . Since $ SE=SF $ , so we get $ \frac{SB^2}{SC^2}=\frac{SF}{CE} \cdot \frac{BF}{SE}=\frac{BF}{CE} $ . ... $ (1) $ Since $ SD $ is $ S - $ symmedian of $ \triangle SBC $ , so we get $ \frac{BD}{DC}=\frac{SB^2}{SC^2} $ . ... $ (2) $ From $ (1) $ and $ (2) $ we get $ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=\frac{CE}{FB} \cdot \frac{BD}{DC}=\frac{SC^2}{SB^2} \cdot \frac{SB^2}{SC^2}=1 $ , so from the converse of Ceva theorem we get $ AD, BE, CF $ are concurrent . Q.E.D That's basically how I did it,except instead of the similarity I used the sine theorem to write $\frac{CE}{FB}$ in terms of $BS$ and $CS$.

Let $EF\cap BC\equiv G$ . It suffices to show that $(G,D;B,C)$ is harmonic .(Let $G$ be closer to $B$) To prove this note that $GS$ is tangent to the circle centered at $P$ . Indeed we have $\angle BSC=90+\angle A$ and $\angle FSE=180-\angle A$ . Thus $\angle FSB + \angle ESC= 90 $ and $FSB$~$SEC$. Also from Menelaus we have $\frac{GB}{GC}=\frac{FB}{EC}=\frac{FB}{ES}\frac{FS}{EC}=(\frac{SB}{SC})^2$ , and our claim is proved . Next we find that $OS$ is the polar of $G$ with respect to the circle centered at $P$. So $D$ is the harmonic conjugate of $G$, as desired.

TelvCohl wrote: My solution: Let $ X=AB \cap (P) $ . Since $ \angle CXA+\angle XAC=\frac{\angle CPB+\angle BOC}{2}=90^{\circ} $ , so we get $ \angle ACX=90^{\circ} $ and $ \angle SBF=\angle SCX=\angle CSE $ , hence $ \triangle SBF \sim \triangle CSE $ . Since $ SE=SF $ , so we get $ \frac{SB^2}{SC^2}=\frac{SF}{CE} \cdot \frac{BF}{SE}=\frac{BF}{CE} $ . ... $ (1) $ Since $ SD $ is $ S - $ symmedian of $ \triangle SBC $ , so we get $ \frac{BD}{DC}=\frac{SB^2}{SC^2} $ . ... $ (2) $ From $ (1) $ and $ (2) $ we get $ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=\frac{CE}{FB} \cdot \frac{BD}{DC}=\frac{SC^2}{SB^2} \cdot \frac{SB^2}{SC^2}=1 $ , so from the converse of Ceva theorem we get $ AD, BE, CF $ are concurrent . Q.E.D Excuse me, why dose $SD$ become the $S-symmedian$ of $\triangle SBC$?

guldam wrote: Excuse me, why dose $SD$ become the $ S- $ symmedian of $\triangle ABC$? That's because $ OB, OC $ are the tangent of $ \odot (SBC) $ , so $ OS \equiv SD $ is the $ S- $ symmedian of $ \triangle SBC $ .

I got it. Thanks!

TelvCohl wrote: My solution: Let $ X=AB \cap (P) $ . Since $ \angle CXA+\angle XAC=\frac{\angle CPB+\angle BOC}{2}=90^{\circ} $ , so we get $ \angle ACX=90^{\circ} $ and $ \angle SBF=\angle SCX=\angle CSE $ , hence $ \triangle SBF \sim \triangle CSE $ . Since $ SE=SF $ , so we get $ \frac{SB^2}{SC^2}=\frac{SF}{CE} \cdot \frac{BF}{SE}=\frac{BF}{CE} $ . ... $ (1) $ Since $ SD $ is $ S - $ symmedian of $ \triangle SBC $ , so we get $ \frac{BD}{DC}=\frac{SB^2}{SC^2} $ . ... $ (2) $ From $ (1) $ and $ (2) $ we get $ \frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=\frac{CE}{FB} \cdot \frac{BD}{DC}=\frac{SC^2}{SB^2} \cdot \frac{SB^2}{SC^2}=1 $ , so from the converse of Ceva theorem we get $ AD, BE, CF $ are concurrent . Q.E.D Can you show me why $ \triangle SBF \sim \triangle CSE $

Panoz93 wrote: Let $EF\cap BC\equiv G$ . It suffices to show that $(G,D;B,C)$ is harmonic .(Let $G$ be closer to $B$) To prove this note that $GS$ is tangent to the circle centered at $P$ . Indeed we have $\angle BSC=90+\angle A$ and $\angle FSE=180-\angle A$ . Thus $\angle FSB + \angle ESC= 90 $ and $FSB$~$SEC$. Also from Menelaus we have $\frac{GB}{GC}=\frac{FB}{EC}=\frac{FB}{ES}\frac{FS}{EC}=(\frac{SB}{SC})^2$ , and our claim is proved . Next we find that $OS$ is the polar of $G$ with respect to the circle centered at $P$. So $D$ is the harmonic conjugate of $G$, as desired. Why is $OS$ the polar of $G$ with respect to the circumcircle of $BSC$?

Stepwise solution coming soon

Attachments:

By Ceva, it suffices to show that $\frac{BD}{DC} = \frac{BF}{EC}$. Since $SD$ is a symmedian in $\triangle SBC$, we get $\frac{BD}{DC} = \frac{SB^2}{SC^2}$. Let $AB\cap (P) = T\neq B$ and let $SE\cap (P) = R$. We have $\angle FBS = \angle SCT = \angle CTR = \angle CSE \implies \triangle BSF \sim \triangle SCE$. So we have $\frac{BF}{BS} = \frac{SE}{SC}$. $(1)$ Also , we have $\frac{CE}{CS} = \frac{SF}{SB} = \frac{SE}{SB}$. $(2)$ Finally dividing $(1)$ and $(2)$ we get $\frac{BF}{CE} = \frac{SB^2}{SC^2} = \frac{BD}{DC}$ as desired.

Notice \begin{align*}\angle SBF+\angle SCE&=\angle ABO+\angle OBS+\angle ACO+\angle OCS\\&=90-\angle C+\tfrac{1}{2}\angle BPS+90-\angle B+\tfrac{1}{2}\angle CPS\\&=\angle A+\tfrac{1}{2}(\angle BPC)\\&=\angle A+\tfrac{1}{2}(180-2\angle A)\\&=90\end{align*}so $\triangle SBF\sim\triangle SCE.$ Also, $(P)$ and $(ABC)$ are orthogonal so $\overline{OS}$ is the $S$-symmedian of $\triangle SBC.$ Hence, $$\frac{AF}{AE}\cdot\frac{EC}{FB}\cdot\frac{DB}{DC}=\frac{EC}{FB}\cdot\frac{SB^2}{SC^2}=\frac{EC}{FB}\cdot\frac{FB\cdot SF}{EC\cdot SE}=1$$and Ceva finishes. $\square$

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