The tangency point lies on $(BXD)$

Apply Menelaus' Theorem on $\triangle ABC$ and transversal $MKL$ to conclude that $\tfrac{BL}{LC}=-\tfrac{BD^2}{CD^2}$ (here lengths are signed). Notice that as $D$ varies with constant velocity on line $BC$, the circumcenter $Z$ of triangle $AXY$ varies on $\ell$ with constant velocity as well. Let the tangents to $\omega$ at $B, C$ meet at $W$. By taking $D=B$ and $D=C$ in the previous argument, we see that $W, D, Z$ are collinear; hence they are always so. Note also that if $E$ lies on $BC$ such that $(BC, DE)=-1$ then $L$ is the midpoint of $DE$. Consider the main Lemma. Line $ZL$ is tangent to $\omega$ for all positions of $D$ on line $BC$. (Proof) Let the tangent from $L$ to $\omega$ touch it at $P$ and meet $\ell$ at point $Z'$. Note that $PD$ and $PE$ are the internal and external bisectors of $\angle BPC$. Consequently, $A, P, E$ are collinear. Let $U=PB \cap AC$ and $V=PC \cap AB$; then line $VW$ passes through $W$ and $Z'$. Also, $UV$ is the polar of $E$ in $\omega$; so $D$ lies on it; hence $Z' \equiv Z$ as desired. $\blacksquare$ From the previous lemma, we conclude that $ZL=LD+ZA,$ so $(AXY)$ and $\mathcal{C}(L, LD)$ are indeed tangent to each other. $\blacksquare$ Note. $\omega$ is the circle $(ABC)$ and $\ell$ the line tangent to it at $A$.

Stats- 3 students solved it in the contest.

Generalization. Let $ABC$ be a triangle inscribed in circle $(O)$. $D$ is an arbitrary point on $BC$. The perpendicular bisectors of $DB,DC$ meet $AB,AC$ at $M,K$, respectively. Line through $D$ and perpendicular to $BC$ cuts $AB,AC$ at $Y,X$. Let point $E$ on $(O)$ such that $AE$ is parallel to $BC$. $ED$ cuts $(O)$ again at $F$. The perpendicular bisector of $DF$ meets $MK$ at $J$. Prove that $(J,JD)$ is tangent to $(AXY).$

The original problem is very nice configuration. Here is my solution. Hope this is not the same with avaliable solution proposed. Let $P$ be the reflection of $D$ through $MK$ then $P$ is on $(L).$ Because $AKDM$ is parallelogram so $AKMP$ is isosceles trapezoid. We have $\angle PMX=\angle PKY$ and $\frac{MP}{MX}=\frac{MD}{MX}=\frac{KD}{KY}=\frac{KP}{KY}.$ Thus, $\triangle PMX\sim\triangle PKY.$ So $\angle PXM=\angle PYK.$ Therefore $P$ is on $(AXY).$ Now we see $(AXY)$ and $(L)$ is tangent because $\angle PAX+\frac{\angle PLD}{2}=\angle XMK+\angle MLD=\angle DBM=\angle XPD.$ The last equality because $\angle BPD=\frac{1}{2}\angle BMD=\angle BXD$ so $BPXD$ is cyclic. We are done.

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We can write this problem for parallelogram as following Let $ABCD$ be a parallelogram. Bisector of $\angle A$ meets $CB,$ $CD$ at $X,$ $Y$, respectively. $Z$ is on $BD$ such that $AZ\perp AX$. Prove that $(Z,ZA)$ is tangent to $(CXY).$ It is nice to compare with problem G3 in IMOSL 2015. I know many extensions for the original problem. But follow me the original problem is nice enough, we not need to see any other generalizaion . Thank Sharygin olympiad to give this nice problem for us !

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buratinogigle wrote: We can write this problem for parallelogram as following Let $ABCD$ be a parallelogram. Bisector of $\angle A$ meets $CB,$ $CD$ at $X,$ $Y$, respectively. $Z$ is on $BD$ such that $AZ\perp AX$. Prove that $(Z,ZA)$ is tangent to $(CXY).$ It is nice to compare with problem G3 in IMOSL 2015. I know many extensions for the original problem. But follow me the original problem is nice enough, we not need to see any other generalizaion . Thank Sharygin olympiad to give this nice problem for us ! Let $T$ be the reflection of $D$ wrt $AC$. Then $T\in (G)$ and because $B$ is the reflection of $D$ wrt midpoint of $AC$, then $TBCA$ is a parallelogram. A well-known result show us that $O$ is the midpoint of arc $ABC$ then $OT=OB$ or $T\in (O).$ Now let $J$ be the intersection of $AC$ and $DE$ then $\frac{JA}{JC}=\frac{DA}{DC}=\frac{TA}{TC}$ then $TJ$ is the bisector of angle $ATC$. This means $\angle JTO=90^\circ$. But $\angle JTG=\angle JDG=90^\circ$ hence $(G)$ is tangent to $(O) $ at $T$. In my opinion, each extension has it own meaning. May be you don't want to see but I would like to But it is true that the original is very beautiful

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An extension. Let $ABC$ be a triangle and $P$ is a point on $BC$. Perpendicular bisector of $PC,$ $PB$ meet $CA,$ $AB$ at $E,$ $F$, respectively. $EF$ meets $BC$ at $D.$ The perpendicular line from $P$ to $BC$ meets $CA,$ $AB$ at $M,$ $N$. $(D)$ is circle center $D$ radius $DP$ and $(K)$ is circumcircle of triangle $AMN$. Prove that $\angle ((D),(K))=|\angle B-\angle C|.$

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Let $I$ be the reflection of $P$ wrt $EF$ then $I\in (D)$. We have $EI=EP=EX=EC$ then $I\in (XPC).$ Similarly , $I\in (PYB)$. Then $I$ is the Miquel point of complete quadrilateral $AXPB.CY.$ This means $I\in (K)$. By angle chasing, $\angle KAC=90^\circ-\angle Y=\angle B $ then $KA$ is tangent to $(O)$. This means $KI$ is also tangent to $(O)$. We have $\angle IOA=2\angle ICA=\angle IEA$ then $A,F,O,E,I,K$ are concyclic. Then $ \angle KID=90^\circ+\angle OIE+\angle EID=90^\circ+\angle OAC+\angle EPC=90^\circ+90^\circ-\angle B+\angle C.$ This means the angle between $(K)$ and $ (D)$ is $|\angle B-\angle C|.$

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We can also see that $(D)$ and $(K)$ intersect at two points, one lying on $(O)$ and the other lying on the circle with diameter $AD$. They will be coincide when $AB=AC.$

Dear friends, I know original problem is very beautiful , but I want to share another generalization for reference Points $M$ and $K$ are chosen on lateral sides $AB,AC$ of an triangle $ABC$ and point $D$ is chosen on $BC$ such that $AMDK$ is a parallelogram. Let the lines $MK$ and $BC$ meet at point $L$. The antiparallel with respect to the sides $BC$ of an angle $BAC$ which passes through $D$ intersects $AB, AC$ at $P, Q$, respectively.Let $X,Y$ be the points respectively on $AB,AC$ such that $DX\perp PC, DY\perp QB$. Prove that the circle with center $L$ and radius $LD$ and the circumcircle of triangle $AXY$ are tangent. I hope my friends like it

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I have another generalization. Given triangle $ABC$ and an arbitrary point $P$ on $BC$. An arbitrary line $d$ through $P$ intersects $AB,AC$ at $Y,X$, respectively. Let $O_1, O_2$ be the circumcenters of triangles $BPY, CPX$, respectively. $O_1O_2$ meets $BC$ at $D$. Let $l$ be line through $P$ and perpendicular to $BC$. Prove that the angle between two circles $(AXY)$ and $(D,DP) $ is $||\angle B-\angle C|-(d,l)|.$ Moreover, $(K)$ and $(D)$ meet at two points, one lying on $(O)$ and the other lying on $(AP).$

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LeVietAn wrote: Dear friends, I know original problem is very beautiful , but I want to share another generalization for reference Points $M$ and $K$ are chosen on lateral sides $AB,AC$ of an triangle $ABC$ and point $D$ is chosen on $BC$ such that $AMDK$ is a parallelogram. Let the lines $MK$ and $BC$ meet at point $L$. The antiparallel with respect to the sides $BC$ of an angle $BAC$ which passes through $D$ intersects $AB, AC$ at $P, Q$, respectively.Let $X,Y$ be the points respectively on $AB,AC$ such that $DX\perp PC, DY\perp QB$. Prove that the circle with center $L$ and radius $LD$ and the circumcircle of triangle $AXY$ are tangent. I hope my friends like it Nice generalization my friend Let $T$ be the reflection of $D$ wrt $MK$. We will show that $(L)$ and $(AXY)$ are tangent at $T$. Let $R$ be the intersection of $BP$ and $CQ.$ We have $A(BCDR)=-1 $ and $AD$ passes through the midpoint of $MK$ then $MK\parallel AR. $ This follows that $MK$ is the midline of triangle $ADR$ or $ T$ lies on $AR$. Since $DT\perp AR$ we get $T$ is the Miquel point of complete quadrilateral $BPCQ.AR.$ Then $\angle DTC=90^\circ-\angle CTR=90^\circ-\angle APB=\angle DYC$. Then $T\in (DYC).$ Simiarly, $T\in (DXB)$. We get $\angle XTY=\angle XTA+\angle YTA=90^\circ-\angle YTD+90^\circ-\angle ABC=180^\circ-\angle ABC-\angle ACB$ then $AXTY$ is cyclic. Trivivaly, $ T\in (L,LD).$ Now two circles are tangent at $T$ iff $\angle YTD=\angle YAT+\angle DLK$. iff $\angle ACB=\angle AKM+\angle L$, that is obviously true.

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A different solution to this beautiful problem: WLOG assume that $D$ is closer to $B$ than $C$. Then $DY \geq DX$ and $LC\geq LB$. By an easy angle chase, one gets that $BM = DM = MX = AK$, $CK = DK = KY = AM$ and $AX = AY$. Now, $BC = ABcosB + ACcosC = 2ABcosB \Rightarrow LC=LB+BC = LB+2ABcosB \Rightarrow LB \cdot LC = LB^2+2AB \cdot LBcosB$ And, $\frac{LB}{LD} = \frac{LM}{LK} = \frac{LD}{LC} \Rightarrow LD^2 = LB \cdot LC = LB^2+2AB \cdot LBcosB$ Also, By cosine rule, $AL^2 = AB^2+LB^2-2AB \cdot LBcos(\angle ABL) = AB^2+LB^2+2AB \cdot LBcosB = AB^2+LD^2$ Now, $\triangle DMX \sim \triangle YAX \Rightarrow \frac{XD}{XY} = \frac{DM}{AY} \Rightarrow BM = DM = AX \cdot \frac{XD}{XY} \Rightarrow AB = 2BM+AX = AX \left(\frac{2XD}{XY}+1 \right)$ $\Rightarrow \sqrt{AL^2-LD^2} \cdot XY = AX(2XD+XY) = AX(XD+YD) \Rightarrow AY \cdot XD+AX \cdot YD-\sqrt{AL^2-LD^2} \cdot XY = 0$ Using the above equation, and by the converse of Casey's theorem on point circles $\odot (A), \odot (X), \odot (Y)$ and circle $\odot (L,LD)$, we get that $\odot (L,LD)$ is tangent to $\odot (AXY)$. NOTE: This solution can also be extended to solve the generalisation given in #13.

Let $Z \equiv \odot(ABC) \cap \odot(AXY)$. $\odot(AXY)$ is orthogonal to $\odot(ABC)$. Note that $Z$ is the centre of spiral similarity sending $\overline{XY} \to \overline{BC}$, thus it is also the centre of spiral similarity sending $\overline{BX} \to \overline{CY}$. Also, $\triangle XDB \sim \triangle YDC$. Therefore, we have \[\frac{ZB}{ZC} = \frac{BX}{CY} = \frac{BD}{CD}.\]Thus, $Z$ lies on the Apollonious circle of $\overline{BC}$ passing through $D$. \[\frac{LB}{LD} = \frac{LM}{LK} = \frac{LD}{LC} \iff LD^2 = LB\cdot LC.\]Thus, $B$ and $C$ are inverses with respect to $\odot(L, LD)$. So, $\odot(L, LD)$ the Apollonious circle of $\overline{BC}$ passing through $D$, and $\odot(L, LD)$ and $\odot(ABC)$ are orthogonal. Thus, $Z$ lies on $\odot(L, LD)$, $\odot(ABC)$ and $\odot(AXY)$, and $\odot(AXY)$ and $\odot(L, LD)$ are orthogonal to $\odot(ABC)$. Thus, $\odot(AXY)$ and $\odot(L, LD)$ are tangent.

anantmudgal09 wrote: Points $M$ and $K$ are chosen on lateral sides $AB,AC$ of an isosceles triangle $ABC$ and point $D$ is chosen on $BC$ such that $AMDK$ is a parallelogram. Let the lines $MK$ and $BC$ meet at point $L$, and let $X,Y$ be the intersection points of $AB,AC$ with the perpendicular line from $D$ to $BC$. Prove that the circle with center $L$ and radius $LD$ and the circumcircle of triangle $AXY$ are tangent. Here is my solution for this problem: Solution

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Let $T$ be reflection of $D$ through $MK$, $U$ $\equiv$ $DX$ $\cap$ $MK$ We see that: $T$ $\in$ ($L$; $LD$) So: $\widehat{LTU}$ = $\widehat{LDU}$ = $90^o$ or $TU$ tangents ($L$; $LD$) at $T$ We have: $\widehat{LTM}$ = $\widehat{LDM}$ = $\widehat{MBD}$, $\widehat{MTK}$ = $\widehat{MDK}$ = $\widehat{MAK}$ Hence: $T$ is the second intersection of ($AKM$) and ($BLM$) or $T$ is Miquel point of completed quarilateral $BCKM.AL$ But: $MB$ = $MT$ = $MD$ = $MX$ so $T$ $\in$ ($BDX$) or $\widehat{BTX}$ = $90^o$ So: $\widehat{ATX}$ = $\widehat{ATB}$ $-$ $\widehat{BTX}$ = $180^o$ $-$ $\widehat{ACB}$ $-$ $90^o$ = $90^o$ $-$ $\widehat{ACB}$ = $\widehat{AYX}$ or $T$ $\in$ ($AXY$) Since: $T$ is the second intersection of ($BDX$) and ($BLM$) so $T$ is Miquel point of completed quarilateral $BDUM.XL$ Hence: $T$ $\in$ ($MUX$) Then: $\widehat{UTX}$ = $\widehat{UMX}$ = $\widehat{UKD}$ = $\widehat{UKT}$ = $\widehat{TAX}$ or $TU$ tangents ($AXY$) at $T$ Therefore: ($L$; $LD$) tangents ($AXY$) at $T$