$(MBD)$ cuts $(O)$ at $G$ => $G$ ∈ $(MEC)$ => ∠$DGE$ = ∠$DAE$ = ∠$DA_1E$ => $G$ ∈ $(DEA_1)$ $GD$ cuts $MC$ at $X$ . We prove that $X$ ∈ $(O)$ <=> ∠$CXG$ = $CBG$. We have : ∠$CBG$ = ∠$ABG$ - ∠$MBA$ - ∠$MBC$ = $180$ - ∠$DMG$ = ∠$EMC$ - ∠$MBA$ = ∠$GMC$ - ∠ $MBA$ ∠$CXG$ = $180$ - ∠$XGC$ - ∠$XCG$ = $180$ - ∠$MBA$ - ∠$MGC$ - ∠$MCG$ = ∠$CMG$ - ∠$MBA$ => ∠$CXG$ = ∠ $CBG$ => $X$∈ $(O)$ $Ge$ cuts $MB$ at $Y$ => $Y$ ∈ $(O)$ ∠$YXG$ = ∠$YXC$ + ∠$CXG$ = ∠$MBC$ +∠$CBG$ = ∠$MBG$ = ∠ $MDG$ => $XY$ // $ED$ => $(GXY )$ tangents to $(GED )$ => $(DEA_1 )$ tangents to $(O)$

my solution )))

Not bad, but we didn't have GeoGebra at the competition. You only had your ruler, compass and your own intuition...

My solution.

let the tangent of $(BOC)$ at $K$ cuts $AB$ at $D$ and $AC $ at $E$ ,the lines $BK$ and $CK$ intersect $(ABC)$ at $B'$ and $C'$ resp. it's easy to notice that $B'C' \parallel DE$ (antiparallel..). Let $L$ the point of intersection of $C'D$ and $B'E$ . Since $K,D,E$ are collinear , applying Pascal's converse to the hexagon $BB'LC'CA $ yields $L$ is on the circumcircle of $ABC$. but $B'C' \parallel DE$ then $(ABC)$ and $(LDE)$ are tangent . in the other hand $\widehat{C'B'L}+\widehat{B'C'L}= \widehat{C'B'C}+\widehat{CB'L}+\widehat{B'C'B}+\widehat{BC'L}= \\ \widehat{C'B'C}+\widehat{B'C'B}+\widehat{BC'L}+\widehat{CB'L}= \pi-2\hat{A}+\hat{A} $ thus $\widehat{DLE}=\hat{A}$ therefore $A'$ is on the circle $(LDE) $ which ends the proof. R HAS

Of course, when the tangency point of the two circles is not stated, it is a good strategy to first find what is the point. So let us define $T$ to be the tangency point of $DE$ and $(BOC)$. Then by Miquel's theorem we obtain $(BDT)$ and $(CET)$ meet again (other than $T$) at $(ABC)$, say $U$. Now we claim that $U$ is our desired tangency point. So this means we shall prove $DEA'U$ is cyclic. This is because $\angle DUE=\angle TBD+\angle TCE=\angle B+\angle C-\angle TBC-\angle TCA=180-\angle A-(180-2\angle A)=\angle A=\angle DA'E$. The last step is to prove both circles are tangent, which amounts to $\angle BUD=\angle DEU-\angle BCU \iff \angle BTD=\angle TCU-\angle BCU=\angle TCB$, which is true since $DT$ is tangent to circle $(BOC)$. This ends the proof. Lastly, it worth noting that this entire problem is the key lemma to the problem 6 of IMO 2011.

navi_09220114 wrote: Of course, when the tangency point of the two circles is not stated, it is a good strategy to first find what is the point. I was doing this for 2 hours before finding it. Let the tangency point be $F$.Let $FB$ cuts $\odot ABC$,$FC$ cuts $\odot ABC$ Thru some angle chasing we have $DE||D^{'}E^{'}$Let $EE^{'}$ cut $\odot ABC$ at $R$.Furthermore let $RD^{'}$ cut $AB$ in $D_{1}$.By Pascals we have $D_{1}$,$F$,$E$ so $D_{1}\equiv DD^{'}$ so $DD^{'}$,$EE^{'}$ are intersected on $\odot ABC$.Let this point be $R$.Again thru some angle chasing $D^{'}E^{'}AB$ is iscosseles trapezium.So R is the center of the homotety that takes circle with radius$\frac{DE}{\sin\alpha}$ that contains $D,E$ to $\odot ABC$ so we are done

Let $DE$ tangent to $\odot(BOC)$ at $P,$ denote by $Q$ the miquel point of $\triangle ABC$ WRT $BCP.$ Line $DQ,EQ$ intersects $\odot(ABC)$ at $X,Y$ respectively. Easy angle-chasing we get $C,X,P$ and $Y,B,P$ are collinear, this is because $\measuredangle YBC=\measuredangle YQC=180^\circ-\measuredangle CQE=180^\circ-\measuredangle CPE=180^\circ-\measuredangle CBP.$ From Reim's Theorem we get $XY\parallel DE.$ Let $F$ be a point on $DE$ such that $FQ$ is tangent to $\odot(ABC).$ Clearly we have \[\begin{aligned} \measuredangle DQF&=\measuredangle DQB-\measuredangle FQB\\ &=\measuredangle DPB-\measuredangle QCB\\ &=\measuredangle PBC-\measuredangle QCB\\ &=\measuredangle QCY=\measuredangle QXY=\measuredangle QED \end{aligned}\]i.e. $FQ$ is tangent to $\odot(DQE)$ as well. From $PO$ is angle bisector of $\angle XPY$ and $OX=OY$, it follows that $O,P,X,Y$ are concyclic $\implies PBY$ and $PXC$ are symmetric over $PO \implies \angle DQE=\angle BAC$ or $180^\circ-\angle BAC.$ i.e. The reflection of $A$ over $DE$ lies on $\odot(DQE),$ as desired. $\square$ [asy][asy] size(9cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt); void b(){ pair A=D("A",dir(52),dir(52)); pair B=D("B",dir(-152),dir(180)); pair C=D("C",dir(-28),dir(0)); pair T=4*B/7+3*C/7; pair O=D("O",origin,dir(45)); pair H=circumcenter(B,O,C); pair P=D("P",IP(circumcircle(B,O,C),L(O,T,-0.01,6)),dir(-135)); pair D=D("D",extension(A,B,P,(H-P)*dir(90)+P),dir(-135)); pair E=D("E",extension(D,P,A,C),dir(-45)); pair Q=D("Q",OP(unitcircle,circumcircle(D,P,B)),dir(-90)); pair F=D("F",extension(E,D,Q,-Q*dir(90)+Q),dir(180)); pair X=D("X",OP(unitcircle,L(D,Q,0,5)),dir(-45)); pair Y=D("Y",IP(unitcircle,L(Q,E,5,-1)),dir(135)); D(unitcircle); D(circumcircle(B,O,C)); D(A--B--C--cycle); D(F--E); D(F--Q,blue); D(D--B); D(E--C); D(D--X); D(E--Y); D(Y--P,dashed+red); D(C--P,dashed+red); D(X--Y,dashed); D(B--Q,dashed); D(C--Q,dashed); D(O--P,dotted); } b(); pathflag=false; b(); [/asy][/asy]

How did you get that this intersection is the tangency point?!

@ K.N $DE$ tangent to $(OBC)$ then $\widehat{DKB}=\widehat{KCB}=\widehat{C'CB}=\widehat{C'B'B}$

navi_09220114 wrote: Of course, when the tangency point of the two circles is not stated, it is a good strategy to first find what is the point. So let us define $T$ to be the tangency point of $DE$ and $(BOC)$. Then by Miquel's theorem we obtain $(BDT)$ and $(CET)$ meet again (other than $T$) at $(ABC)$, say $U$. Now we claim that $U$ is our desired tangency point. So this means we shall prove $DEA'U$ is cyclic. This is because $\angle DUE=\angle TBD+\angle TCE=\angle B+\angle C-\angle TBC-\angle TCA=180-\angle A-(180-2\angle A)=\angle A=\angle DA'E$. The last step is to prove both circles are tangent, which amounts to $\angle BUD=\angle DEU-\angle BCU \iff \angle BTD=\angle TCU-\angle BCU=\angle TCB$, which is true since $DT$ is tangent to circle $(BOC)$. This ends the proof. Lastly, it worth noting that this entire problem is the key lemma to the problem 6 of IMO 2011. Could you please explain how have you used Miquel theorem ? Cause I can't see it , thanks .

he used miquel for triangle ADE.

Cute problem! Let $T=(BDP) \cap (CEP) \cap (ABC)$ be the Miquel point of $\triangle BPC$ with respect to $\triangle ADE$. We claim that $T$ is the desired tangency point. Note that $\measuredangle{ETD}=\measuredangle{ETP}+\measuredangle{PTD}=\measuredangle{ACP}+\measuredangle{PBA}=\measuredangle{CAB}+\measuredangle{PBC}=\measuredangle{BAC}=\measuredangle{EA'D}$, thus $T \in (A'DE)$. Suppose, for the sake of contradiction, that there exists $T' \neq T$ lying on both $(ABC)$ and $(A'DE)$. Let $P'=(BDT') \cap (CET')$. By Miquel, $P' \in DE$. Furthermore, $\measuredangle{BP'C}=\measuredangle{BDT'}+\measuredangle{T'EC}=\measuredangle{ADT'}+\measuredangle{T'EA}=2\measuredangle{BAC}=\measuredangle{BOC}$, therefore $P' \in (BOC)$, implying $P'=P$ and $T'=T$, a contradiction.

Let $X$ be the second intersection of $(BDP)$ and $(CEP)$, where $P$ is the tangency point of $DE$ and $(BOC)$. Since $$\angle BXC = \angle BXP + \angle PXC = \angle ADE + \angle AED = 180^\circ -\angle A$$, we have that $X\in (ABC)$ as well. Our objective is to show that $X$ is the desired tangency point. Now, we claim that $DEA'X$ is cyclic. This is true since \begin{align*}\angle DXE = \angle DXP + \angle PXE = \angle DBP + \angle PCE = \angle ADE + \angle AED- (\angle DPB + \angle EPC) \\ = 180^\circ-\angle A - (180^\circ-\angle BOC) = \angle A=\angle DA'E.\end{align*} Finally, let $XD$ and $XE$ intersect the circumcircle of $\triangle ABC$ again at $F$ and $G$, respectively. We also have that $\angle BXD =\angle BPD = \angle BCP$, so $C$, $P$, and $F$ are collinear. Analogously, $B$, $P$, and $G$ are collinear. To finish, we have that \[\angle FGX = \angle FCX = \angle PCX = \angle PEX\]so $FG\parallel DE$. By the converse of Reim's theorem, this implies $(XDE)$ and $(XFG)$ are tangent at $X$, as desired.

Let $DE$ be tangent to $BOC$ at $P$. Consider $B' = BP \cap (ABC)$, $C' = CP \cap (ABC)$. Note that $OP$ bisects $\angle{BPC}$. so $B'C'$ is the reflection of $CB$ over $OP$. In particular, $B'C'$ subtends an arc of angle $\angle{A}$. Now, note that by Reim, $B'C'$ is parallel to $PP = DE$. Moreover, by Reverse Pascal, $B'E \cap C'D = F$ is on $(ABC)$. Now, $\angle{EFD} = \angle{A}$, so thus $F \in (A'DE)$, but now since $DE // B'C'$, $(A'FDE)$ is tangent to $(AB'C')$. This thus proves the problem.

Let $P$ be the tangency point of $DE$ and $(BOC)$ Let $F=(BDP)\cap (CEP)$ We will prove that $F$ is the desired tangency point Since $\angle BFC=\angle BFP+\angle CFP=\angle ADE+\angle AED=180-\angle BAC$ we have that $F$ lies on $(ABC)$ $\angle DFE=\angle DFP+\angle PFE=\angle DBP+\angle PCE=(\angle ADE-\angle DPB)+(\angle AED-\angle EPC)=(\angle ADE+\angle AED)-(\angle DPB+\angle EPC)=(180-\angle BAC)-(\angle PCB+\angle PBC)=(180-\angle BAC)-(180-\angle BPC)=\angle BPC-\angle BAC=2\cdot \angle BAC-\angle BAC=\angle BAC$ So $F$ lies on $(DA'E)$ It is left to show that $(DEF)$ and $(ABC)$ are tangent Let $FD, FP, FE$ intersect $(ABC)$ at $G, I, H$ $\angle CAI=\angle CFI=\angle CFP=\angle AEP$ so $AI\parallel DE$ $\angle AHG=\angle AFG=\angle AFB-\angle DFB=\angle ACB-\angle DPB=\angle ACB-\angle PCB=\angle ECP=\angle EFP=\angle HFI=\angle HAI$ Hence $AI\parallel GH$ This means that $DE\parallel GH$ which means that $\Delta FDE$ and $\Delta FGH$ are homothetic which implies that $(DEF)$ and $(ABC)$ are tangent as desired

$M$ is the tangnet point of $DE$ to $(BOC)$. Let $(BDM)$ intersect $(ABC)$ at $X$ . $\measuredangle EMX = 180^{\circ}-\measuredangle DMX$ and $\measuredangle ECX = \measuredangle DMX$, thus we get $EMXC$ as cyclic. Claim- $DEXA'$ is cyclic. Proof- $\measuredangle EXD = \measuredangle EXM +\measuredangle MXD = \measuredangle ECM + \measuredangle MBD = \measuredangle ECB-\measuredangle BCM + \measuredangle DBC - \measuredangle CBM = (\measuredangle { 180^{\circ} -BAC}) - (\measuredangle {180^{\circ} - CMB}) = \measuredangle BAC= \measuredangle BA'C$. Now, let $GX$ be tangent to $(ABC)$, Claim- $GX$ is also tangent to $(A'ED)$. Proof- $\measuredangle EXG =\measuredangle EXC + \measuredangle CXG = \measuredangle EMC + \measuredangle CBX =\measuredangle MBC+\measuredangle CBX = \measuredangle MBX = \measuredangle MDX = \measuredangle EDX$ Thus, it finishes the proof that $(ABC)$ and $(A'DE)$ are tangent at $X$.