Let $ N \equiv {} AO \cap BC,$ $V \equiv{} AO \cap EF$ and $Q \equiv{} EF \cap BC.$ Let $ T$ be projection of $ C$ on $ AN$ and $ P \equiv{} AQ \cap CT.$ $ \angle CAD = \angle CBA$ implies that $ EF \parallel AC.$ Therefore, $ \triangle ATC$ and $\triangle VZQ$ are similar (Z is the projection of Q on AN). By Menelaus' theorem for $\triangle ANQ$ cut by the transversal $\overline{TPC},$ we obtain: $\frac {AP}{PQ} = \frac {CN}{CQ} \cdot \frac{TA}{TN}.$ But $ \frac {CN}{NQ} = \frac {AN}{AV},$ due to $ VQ \parallel AC.$ On the other hand, $ \frac {AN}{AV} = \frac {ZN}{ZV}$ (A,V,Z,N are harmonically separeted). $ \frac {AP}{PQ} = \frac {ZN}{ZV} \cdot \frac{TA}{TN}.$ But $ \frac {TA}{ZV} = \frac {TC}{ZQ} = \frac {TN}{ZN}$ $\Longrightarrow$ $ AP = PQ$ $\Longrightarrow$ $ P$ is the midpoint of $ AQ.$ Consequently, $ T$ is the midpoint of $ AZ$ $\Longrightarrow$ $ AT$ passes through $ G$ and the conclusion follows.

A more synthetic approach: Notice that from the hypothesis we know that $ CA$ is tangent to the circumcircle $ \Gamma$ of $ ABD$. Consider the point $ A$ as a degenerated circle, and thus, since the radical axis of $ A$, $ \Gamma$ and $ (BDEF)$ are concurrent at the circles' radical center, we deduce that $ C$ is the radical center of the three circles. Thus, we are left to show that $ M$ lies on the radical axis of $ A$ and $ (BDEF)$. But this is immediate, since $ G$ lies on the polar of $ A$ wrt. $ (BDEF)$ and so, if we denote by $ X$, $ Y$ the intersections of $ AG$ with $ (BDEF)$, we have that $ XA / XG = YA / YG$, which is equivalent with $ MG^{2} = MA^{2} = MX \cdot MY$. In conclusion, $ M$ lies on $ AO$ since $ MA^{2}$ is the power of $ M$ wrt. $ A$ and $ MX \cdot MY$ the power of $ M$ wrt. $ BDEF$.

It might be interesting to formulate the problem this way: Given a cyclic quadrilateral $ BDFE$. Rays $ BD$ and $ EF$ meet at $ R$, and rays $ BE$ and $ DF$ meet at $ A$. Let a line through $ A$ parallel to $ EF$ meet $ BD$ at $ C$. Let $ FB$ and $ ED$ intersect at $ G$, and $ GR$ and $ AC$ intersect at $ L$. Show that $ AC = CL$. Would this inspire a different approach?

hollandman wrote: It might be interesting to formulate the problem this way: Given a cyclic quadrilateral $ BDFE$. Rays $ BD$ and $ EF$ meet at $ R$, and rays $ BE$ and $ DF$ meet at $ A$. Let a line through $ A$ parallel to $ EF$ meet $ BD$ at $ C$. Let $ FB$ and $ ED$ intersect at $ G$, and $ GR$ and $ AC$ intersect at $ L$. Show that $ AC = CL$. Would this inspire a different approach? The pencil $ R.BAFG,$ is in harmonic conjugation as well. So, because of $ AL\parallel RF,$ we conclude that $ AC = CL$ $ ,(1)$ Similarly, we have that $ AZ = AN$ $ ,(2)$ where $ Z$ is the point of intersection of the line segment $ BF,$ from the line through $ A$ and parallel to $ DE$ and $ N\equiv GR\cap AZ.$ From $ (1),$ $ (2)$ $ \Longrightarrow$ $ CZ\parallel GR$ $ ,(3)$ But, it is well known that $ OA\perp GR$ $ ,(4)$ $ ($ The line connecting the circumcenter of a cyclic quadrilateral, with the point of intersection of its diagonals, is perpendicular to the line connecting the points of intersection of its opposite sidelines. $ )$ Hence, from $ (3),$ $ (4),$ we conclude that $ OA\perp CM$ and the proof of the proposed problem is completed. REFERENCE. - http://www.mathlinks.ro/Forum/viewtopic.php?t=110887 Kostas Vittas.

Call the intersection of $ BC$ and $ EF$ is $ P$, $ PG$ and $ AD, AC$ are $ X$ and $ T$. We know that $ G$ is the orthorcenter of $ \Delta ABC$, and $ (AXDF) = - 1$, $ AT//PE$ hence $ CA = CT$. But $ PG \perp AO$, so $ AO \perp CM$. An old idea for the problem!

Dear Fang-jh, Cosmin and Mathlinkers, can some one draw the figure... I don't arrive to the conclusion which is asked. Thank you Sincerely Jean-Louis

Dear Mathlinkers, pleasr can some one draw the figure of the initial, problem... Is there a typo in the text of the problem? Sincerely Jean-Louis

If t is tangent to the circumcircle (P) of triangle ABC at A and t' reflection of t in AC, then t' cuts BC at D, such that <CAD = <ABC = -<CBA (not <CAD = <CBA).

Dear Yetti, now it is clear for me... because I was taking the synthetic result of Cosmin... Thank you very much Sincerely Jean-Louis

Here is my solution (using Inversion:D): Let $ P$ be the intersection of $ EF$ and $ BC$. Then, it is very well- known that $ G$ is the orthocenter of triangle $ OPA$ (This result can be proved by using Pole-Polar Theory). Let $ A_1$, $ P_1$, $ O_1$ be the intersections of $ (AG,OP)$, $ (PG,OA)$, $ (OG,AP)$, respectively. Consider the inversion through pole $ O$, power $ R^2$, where $ R$ is the radius of $ (O)$, we get, $ I(O,R): B\mapsto B$, $ D\mapsto D$, $ E\mapsto E$, $ F\mapsto F$. Therefore, $ I(O,R): BD\mapsto (OBC)$. Hence, $ P\mapsto A_1$ and $ A_1 \in (OBC)$. Now, it is easy to notice that $ \overline {PD}. \overline {PB} = \overline {PA_1}. \overline {PO} = \overline {PG}. \overline {PP_1}$, which implies $ BP_1GD$ is concyclic quadrilateral, let $ (I)$ be the circumcircle of $ BP_1GD$. $ I(O,R): G\mapsto O_1$, $ P_1\mapsto A$, $ B\mapsto B$, $ D\mapsto D$. Therefore, $ I(O,R): (I)\mapsto (ABD)$. In the other hand, $ I(O,R): C\mapsto C'$, then $ C'\in (OBC)$, hence, $ I(O,R): CP_1\mapsto (OAC')$. Now, it is easy to notice that $ CA$ is the tangent of $ (ABD)$, which implies that $ CA^2 = \overline {CD}.\overline {CB} = \overline {CC'}.\overline {CO}$, which also implies that $ CA$ is also tangent of $ (AOC')$. Therefore, $ (ADB)$ internally tangents $ (AOC')$. Therefore, $ CP_1$ is also a tangent of $ (I)$. Therefore, $ CP_1^2 = \overline {CD}. \overline {CB} = CA^2$, hence $ CP_1 = CA$, which implies that $ C$ lies on the bisector of $ AP_1$, $ M$ also lies on the bisector of $ AP_1$ (Because $ M$ is considered the circumcircle of $ (AP_1G)$). Therefore, $ CM$ is the bisector of $ AP_1$, hence, $ CM\bot AO$. Our proof is completed

Sorry to revive this old thread but I have an alternative proof to this wonderfull problem. In my proof, $Q$ is the midpoint of $AG$. [geogebra]623d3e5ef774564f959c2dd8f96536adbf98793b[/geogebra]

Fang-jh wrote: Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD = \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$ When $ D$ lies on its sideline $ BC$ then D lies on between B and C or not ?

Please reply me ! I isn't good at English.

Nice problem. By Brokard's Theorem, and angle chasing it suffices to showing that if $AG$ meets $EF,BC$ at $P,T$ respectively then observe that $(A,G;P,T)=-1$ and so $TP.TM=TG.TA$ and the angle condition means $EF$ parallel to $CA$ and we want $CM$ parallel to $HG$. Now this is trivial by converting these angles into ratios.

[asy][asy] import graph;import geometry;import olympiad;size(10cm); pair A=(6,6),B=(0,0),C=(10,0),Dee,E,F,G,O,M,X,Y,Z; real beta=degrees(C/A);pair Dprime=rotate(beta,A)*C; Dee=extension(A,Dprime,B,C); E=0.65*A; path p=circumcircle(B,Dee,E); pair Ghost=E+C-A; F=extension(E,Ghost,A,Dee); G=extension(B,F,Dee,E); M=(A+G)/2; O=circumcenter(B,Dee,E); X=extension(E,F,B,C);Z=extension(X,G,A,B);Y=extension(X,G,A,C); draw(A--B--C--A--Dee,blue);draw(E--X,magenta);draw(E--Dee,magenta);draw(B--F,magenta);draw(Z--Y--C--M,heavycyan);draw(A--X,heavycyan);draw(A--G,magenta);draw(A--O,heavygreen+dotted);draw(p,orange); dot(A);dot(B);dot(C);dot(Dee);dot(E);dot(F);dot(G);dot(O);dot(M);dot(X);dot(Y);dot(Z); label("$A$",A,N);label("$B$",B,SW);label("$C$",C,SW);label("$D$",Dee,SE);label("$E$",E,N);label("$F$",F,ENE);label("$G$",G,SSW);label("$O$",O,W);label("$M$",M,NW*0.4);label("$X$",X,NE);label("$Y$",Y,S);label("$Z$",Z,WNW); [/asy][/asy] Suppose $EF\cap BC=X,XG\cap AC=Y,XG\cap AB=Z$. Note that $\angle FAC=\angle DAC=\angle ABC=\angle EBD=$ $\angle EFA$, so that $FE||AC$; let $P_{\infty}$ be the point at infinity in this direction. In the triangle $XBE$, $XZ,ED,BF$ are concurrent cevians and $DF\cap BE=A$, so we have $(E,B;A,Z)-1\implies X(E,B;A,Z)=-1$. Intersecting this pencil with $AC$ we have $(A,Y;C,P_{\infty})=-1\implies C$ is the midpoint of $AY$. But $M$ is the midpoint of $AG$, so $CM||YG\implies CM||XG...(\star )$. But applying $\textsc{Brokard's Theorem}$ to the cyclic quadrilateral $DFEB$, we $XG$ is the polar of $A$ w.r.t. $(O)\implies XG\perp AO$. Combining this with $(\star )$, we arrive at $CM\perp AO$, as desired. $\blacksquare$

This ends up being a rather nice usage of the perpendicularity lemma and some length chasing. Let $r$ be the circumradius of $(O)$. We want to show, by the perpendicularity lemma, that \[ CA^2 - OC^2 = MA^2 - MO^2 \] Observe that since $CAD \sim CBA$, we have that $CA^2 = CD(CB)$. Then, since $CD(CB) = \operatorname{Pow}(C, (O)) = CO^2 - r^2$, we have that $CA^2 - OC^2 = -r^2$. Now I claim that $MA^2 - MO^2 = -r^2$, which will of course finish the problem. By the parallelogram law, we can compute \[ MO^2 - AM^2 = \frac{2OG^2 + 2OA^2 - 2AG^2}{4} = \frac{2(OG)(OA)\cos{\angle{GOA}}}{2} = OG(OA)\cos{\angle{GOA}} \] Now, we're basically done: $OA\cos{\angle{GOA}}$ is the length of the projection from $G$ onto $OA$, and this rearranges to $G$ lying on the polar of $A$ wrt $(O)$, which is true by Brokard's theorem.

My solution : $EF$ cuts $BD$ at $T$, $GT$ cuts $AB$ at $H$ , cuts $AC$ at $A'$ We have : $(HE<HA) = -1 \implies T(BE,HA) = -1 \implies T(CE,A'A) = -1$ $(1)$ But we have $\angle AEF = \angle ADB = \angle CAB$ so $AA' \parallel EF$. $(2)$ Combine $(1)$ with $(2)$ we have $C$ is the midpoint of $AA'$ $\implies CM \parallel TG \perp AO$

We prove the statement by showing that points $C$ and $M$ lie on the radical axis of the degenerate circle $A$ and circle $(O)$. It is clear that $C$ lies on the circle since $CA^2 =CD \cdot CB$ is given in the statement. For $M$, $A$ and $G$ are conjugates wrt $(O)$, so the circle $O$ is orthogonal with the circle having $(AG)$ as its diameter. This is enough to finish the problem. $\blacksquare$

. INSIGHT

.Well , let say $T\equiv FE\cap BC$,after applying brocard we see that $TG$ is perpendicular to $AO$.Now we have to prove that $CM$ is perpendicular to $AO$.So, the problem turn to showing that $TG$ is parallel to $CM$. PROOF

China TST 2009 P1 wrote: Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD = \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$ Solution: Observe $\angle CAD= \angle ABD =\angle AFE$ $\implies$ $AC||EF$. Let $EF$ $\cap$ $BC$ $=$ $X$. Brokard's Theorem implies, $GX$ $\perp$ $AO$. Hence we want to show $GX$ $||$ $CM$. \begin{align*} -1= (A, G ; AG \cap EF, AG \cap BC) \overset{X}{=} (A, XG \cap AC; \infty_{AC}, C) \end{align*}Hence, $CM || XG$ as desired $\qquad \blacksquare$

Top 10 projective geo lets go! By anti-parallels we have that $EF \parallel AC$, let $EF \cap BC=J$, $AJ \cap ED=X$ and $A'$ reflection of $A$ over $C$. Now by Ceva-Menelaus config in $\triangle BJA$ and projecting we get: $$-1=(D, E; X, G) \overset{J}{=} (C, \infty_{AC}; A, GJ \cap AC) \implies GJ \cap AC=A' \implies GJ \parallel MC$$Now by brokard theorem we get $AO \perp GJ \parallel MC$ thus we are done.