We can write that pro like that: triangle $DAX$, $O$ is the midpoint of $AX$,$E$ is the orthocenter of $ADX$, $AE,XE$ meet $DX,DA$ at $C,B$. $F$ is the intersection point of $DE$ and the circle with diameter $AX$. And easy to see that : $\angle XFC=\angle XAC=\angle XDE$ and $\angle XFE=\angle XAF=\angle XBF$ and we have $XF$ is the tangent common of $(CDF)$ and $(BEF)$ we are done.

Nice and easy one. Let $O_1$ be the circumcenter of $BFE$ and $O_2$ be the circumcenter of $DFC$ It suffices to prove that $O_1,F,O_2$ are collinear. Let $DC$ meets the circle at $S$. Then E is the orthocenter of DAS, so $S,B,E$ are collinear. Then we also have that $\angle{O_1FE}=90^{\circ}-\angle{EBF}=\angle{FCA}$ And $\angle{FCO_2}=90^{\circ}-\angle{FDC}=\angle{FEC}$ So $\angle{O_1FE}+\angle{FCO_2}+\angle{EFC}=180^{\circ}$ and we are done.

This is the easiest geometry problem in shortlist. I think the geometry problem may be chosen so that it is more challenging. Anyway, this is a nice problem.

It's easy to prove why we can reword the problem as follows: Let acute-angled triangle $ABC$ have orthocentre $H$ and orthic triangle $DEF$. Suppose $BH$ meets the circle going through the points $A,C,D$ and $F$ at $P$. Show that $(HFP)$ and $(BPD)$ are tangent at $P$. To do this, we shall show $CP$ is a common tangent to the two circles. Note that $HDBF$ is cyclic due to the right angles. Thus $CH\cdot CF=CD\cdot CB$. Now, an angle chase shows that $\angle CHB=180^{\circ}-A=\angle FPC$ so it follows that triangles $CHP$ and $CPF$ are similar. Then $\frac{CP}{CH}=\frac{CF}{CP}$, which implies $CP^2=CH\cdot CF=CD\cdot CB$. It then follows by power of a point that $CP$ is tangent to $(HFP)$ and $(BPD)$, so the result follows.

It is sufficient to prove that <FBE+<FDC =<EFC Let the intersection point of DE and AO be point X, then we have that AXCD is cyclic, from where we get that <EDC=<EAX=a Let the intersection point of AO and CD be point Y, then we have that Y lies on C(O,OA) (because <ACY=90 and AY is the diameter of this circle). <AYD=90-a=<DBC=<DEC. Hence BECD is cyclic and <EBC=<EDC=a. If <EBF=b, then <CBF=<FAC=b-a. <AFC=<ABC=180-<DBC=90+a. <ACF=180-<FAC-<AFC=90-b, hence <EFC=180-<DEC-<FCA=a+b=<FBE+FDC

Similar with the nice WakeUp's proof ==> See PP3 from here.

A proof by inversion: $DC$ cuts again the circle at $N| AD$ is a diameter of the circle $\odot (ABC)$, $DE$ cuts $AN$ at $P$ and the circle at $F, M$. Inversion of pole $D$ and power $AD\cdot BD$ sends the circle $\odot (DCEB)$ to the line $AN$, hence $E$ to $P$, $C-N, F-M, B-A$, hence the circle $\odot (DCF)$ to the line $MN$, and the circle $\odot (BEF)$ to the circle $\odot (AMP)$, which is tangent to $MN$, done. Best regards, sunken rock

Let $AO \cap \Gamma=G$ $\angle ACD= \angle ACG=90$ So $D,B,C$ are colinear . Note that $E$ is the orthocenter of $\triangle ADG $ . As $\angle ABG =90$ , $G,E,B$ are colinear . Draw a line $DH||BG$ so that $FB \cap DH=H$ Now $\angle DHF=\angle DGF=\angle GFC $ . So $C,D,H,F$ are concyclic . So circumcircle of $\triangle FEB ,\triangle FDH $ has a common tangent at $F$ . Which imply they are tangent at $F$

With the following two statements I have some problem. Probably the image is different bacteriam wrote: ......................... Now $\angle DHF=\angle DGF=\angle GFC $ . So circumcircle of $\triangle FEB ,\triangle FDH $ has a common tangent at $F$ . Which imply they are tangent at $F$ But with this I agree , but with a different proof ! bacteriam wrote: ......................... ....,so $C,D,H,F$ are concyclic . I think , I need your help.

Dear Mathlinkers, we can also think to a converse of the pivot theorem applied to the triangle HBC. Sincerely Jean-Louis

Following a challenge from school5, see below: Easily $E$ is the orthocenter of $\Delta ADN$, hence $B-E-N$ are collinear. We shall use inversion of pole $F$ and a random power. Let $X'$ be the image of $X$; we need to prove $C'D'\parallel B'E'\ (\ *\ )$. We have $DCC'D', BEE'B', BFCN$ cyclic, hence $\angle C'D'F=\angle DCF=\angle FBN=\angle FE'B'$, so $\angle C'D'F=\angle FE'B'$ and $(*)$ is true. Best regards, sunken rock

It is enough to show that $ \angle FBE= \angle FCD, \angle ABE+\angle FBE+90 - \angle FCD=180 $ so we have to prove that $ \angle ABE=90 $ Let $ AO $ intersects the circle at $ S $. Since $ \angle ACD= \angle ASC=90, S,C,D $ are collinear, and note that $ E $ is the orthocenter in $ ADS $, so $ SE\perp AD $ but $ SB\perp AD $ which leads to $ EB\perp AD $ , QED

Can someone explain me : l"et D be the point of the intersection of the line AB and the perpendicular line to AC at C"? Because in my pic iy is not tangent

Let $K$ be diametrically opposite $A$ on $\Gamma$. Then $KB \perp AB \perp BE, KC \perp AC \perp CD$ so $KEB, KCD$ are straight lines. It follows $E$ is the orthocenter of $\triangle ADK$. Hence $CEBD$ is cyclic, so $KE \cdot KB = KC \cdot KD$, by intersecting chords. Let $F' \equiv FE \cap \Gamma$. Then $F'$ is the reflection of $F$ in $AK$, so $FK = KF'$. \begin{align*}\therefore \angle KEF = 180^{\circ} - (\angle F'FK +\angle BKF) = 180^{\circ} - (\angle KAF + \angle BAF) =180^{\circ} - \angle A = \angle BFK. \end{align*} Hence $\triangle KEF \sim \triangle KFB$, so $\frac{KF}{KE} = \frac{KB}{KF} \implies KF^2 = KE \cdot KB = KC \cdot KD$, so $KF$ is tangent to circles $BFE, CFD$, from the converse of the tangent-secant theorem. It follows the circles are tangent at $F$.

This problem is very easy even for #1. $\angle{OAC}=\theta \implies \angle{AOC}=180-2\theta \implies \angle{CBD}=90-\theta=\angle{DEC} \implies$ points $B,D,C,E$ are concyclic.This yeilds $\angle{EBD}=\angle{DCE}=90^{\circ}$.Also note that $\angle{DBF}=\angle{FCA} \implies \angle{EBF}=\angle{FCD}$ or the circumcircles of $EBF,FCD$ are tangent at $F$ as required.

My solution. let $X$ be the intersection of $DE$ with $AO$, and let $Y$ be the intersection of $AO$ with $(\Gamma).$ since, $C$ lies on $(\Gamma).$ and, the quadrilateral $DCXA$ is Cyclic ($\measuredangle DCA=\measuredangle DXA=90^\circ$). we have, $\measuredangle DCA+\measuredangle ACY=90^\circ+90^\circ=180^\circ.$ and thus, the points $D$,$C$,$Y$ are collinear. and since, $\measuredangle DCE=\measuredangle YXE=90^\circ$ we deduce that the quadrilateral $ECYX$ is Cyclic. Or, $\overline{DE} \cdot \overline{DX}=\overline{DC} \cdot \overline{DY}=\mathcal{P}_{(\Gamma)}(D)=\overline{DB} \cdot \overline{DA}.$ Therefore, the quadrilateral $DCEB$ is also Cyclic. now, it is easy to see that, $\measuredangle DFB=\measuredangle CAB+(\measuredangle FDC+\measuredangle DCF)=(\measuredangle CAB+\measuredangle XAC)+\measuredangle DCF=\measuredangle DEB+\measuredangle DCF.$ this prove that the circumcircles of triangles $BFE$ and $CFD$ are tangent at $F$.

cefer wrote: Let $A$, $B$ and $C$ be points lying on a circle $\Gamma$ with centre $O$. Assume that $\angle ABC > 90$. Let $D$ be the point of intersection of the line $AB$ with the line perpendicular to $AC$ at $C$. Let $l$ be the line through $D$ which is perpendicular to $AO$. Let $E$ be the point of intersection of $l$ with the line $AC$, and let $F$ be the point of intersection of $\Gamma$ with $l$ that lies between $D$ and $E$. Prove that the circumcircles of triangles $BFE$ and $CFD$ are tangent at $F$. Let $\overline{DE}$ meet $\Gamma$ again at $L$ and line $\overline{AO}$ at $K$. Suppose $\overline{AO}$ meets $\Gamma$ again at $R$. Note that $R$ lies on line $\overline{CD}$ and $E,C,R,K$ are concyclic. By Power of Point, $$\overline{DA} \cdot \overline{DB}=\overline{DC} \cdot \overline{DR}=\overline{DE} \cdot \overline{DK}$$hence $A,B,E,K$ are concyclic proving $\angle ABE=90^{\circ}$. Note that points $F$ and $L$ are symmetric in line $\overline{RA}$. Accordingly, $$\angle BFD=\overarc{BL}=\overarc{BR}+\overarc{RL}=\overarc{BR}+\overarc{FR}=\angle BAK+\angle FLR=\angle BED+\angle FCD,$$proving $\odot(BFE)$ and $\odot(CFD)$ are tangent at point $F$ (all arcs taken with respect to $\Gamma$). $\blacksquare$

So this is problem 1. BMO 2012. but if we look at JBMO shortlist 2009. we see that it's G5.. Any explanation?

Let $A'$ be the A-antipode. By noting that $\angle ABA' = 90$ and $\angle ACA' = 90$, we get that $E$ is the orthocentre of $ADA'$, and that $AC$, $A'B$, and $DP$ are the altitudes, where $P = \lambda \cup \overline{AA'}$. Now it's well-known that $D$ is the radical centre of $ABEP$, $ABCA'$, $ECA'P$ and hence inversion centre $D$ radius $sqrt{DB \cdot DA}$ sends $A \leftrightarrow B$, $E \leftrightarrow P$, $C \leftrightarrow A'$, and $F \leftrightarrow G$, where $G$ is the other point of intersection of $\lambda$ with $\Gamma$. It suffices to show that $(B'E'F')$ is tangent to $C'F'$ at $F'$. Yet $\angle B'E'F' = 90$ by $\lambda \perp B'C'$, which equals $\angle C'F'B'$ by Thales. Hence, by converse of alternate segment, done. $\blacksquare$

To prove the result, we claim $C'F$ is tangent to both circles. Denote $C'$ as the antipode of $A$, and $T$ as the foot of the altitude from $D$ to $AC'$. Since $\angle ACC' = 90$, we have $C'CD$ collinear, and $E$ is the orthocenter of $AC'D$. We use the converse of Power of a Point to get \[C'F^2 = C'T \cdot C'A = C'C \cdot C'D = C'E \cdot CB.\] Thus $C'F$, $(BFE)$, and $(CFD)$ are all tangent at $F$. $\blacksquare$ [asy][asy] size(300); pair D, A, C1, B, C, E, F, O; D = dir(70); A = dir(220); C1 = dir(320); B = foot(C1, A, D); C = foot(A, C1, D); E = extension(A, C, B, C1); F = IP(E--D, circumcircle(A, B, C)); O = .5*A + .5*C1; draw(D--C1--A--D--foot(D, A, C1)^^B--C1--E--A--C^^C1--F--A); draw(circumcircle(B, E, F)^^circumcircle(C, F, D)); draw(circumcircle(A, B, C), blue); markscalefactor = .008; draw(rightanglemark(A, F, C1)^^rightanglemark(D, foot(D, A, C1), A)^^rightanglemark(D, C, A)^^rightanglemark(C1, B, D)); dot("$D$", D, NW); dot("$A$", A, W); dot("$C'$", C1, E); dot("$B$", B, NW); dot("$C$", C, NE); dot("$E$", E, 2dir(295)); dot("$F$", F, dir(40)); dot("$O$", O, S); dot("$T$", foot(D, A, C1), S); [/asy][/asy]