My solution: From the condition we get $ \frac{AB}{BC}=\frac{AD}{DC} $ . ie. $ ABCD $ is a harmonic quadrilateral Since $ \angle CBM=\angle DBA $ , so we get $ \angle AMB=\angle ACB+\angle CBM=\angle ADB+\angle DBA=\angle DCB $ . Similarly, we can prove $ \angle DMA=\angle DCB $ . Since $ \angle CME=\angle DCB=\angle CDE $ , so we get $ C, D, E, M $ are concyclic . Since $ \angle DEC=\angle DMA=\angle DCB=\angle DFB $ , so we get $ BF \parallel CE $ and $ BCEF $ is a parallelogram . Q.E.D

I have a similar solution: $\widehat{DEM}=\widehat{EBC}=\widehat{ABD}=\widehat{ACD}\Rightarrow D,M,C,E$ concyclic. $\widehat{DEC}=\widehat{DEM}+\widehat{MEC}=\widehat{ABD}+\widehat{DAB}=\pi-\widehat{DAB}=\widehat{DCB}=\widehat{FBC}$, hence the conclusion ($FB \parallel EC$)

Alternative: let the line parallel with $AC$ which passes through $D$ cut the circumcircle of $ABCD$ in $D'$. Then $BD'$ goes through the midpoint of $CD$ (Pascal on $D'DFACB$, since $BC // DF$ and $DD'//AC$) and through $M$ ( $BD$ and $BM'$ are isogonally conjugate) Hence the diagonals of $CBFE$ goes through one midpoint and one pair of sides is parallel, so $BCEF$ is a parallellogram.

Other solution: It is enough to prove that $CE \parallel BF$, since we already have $FE \parallel BC$. First, how $\dfrac{BA}{BC} = \dfrac{DA}{DC}$, we have that $ABCD$ is harmonic, thus the tangents from $B$, $D$ and the straight line $AC$ are concorrent in a point $P$. Now, let $R$ be the intersection of $BD$ and $AC$. Since $BD$ is the polar of $P$, $P, R, C, A$ is a harmonic bundle, i.e., $(P, R; C, A) = -1$. But since $M$ is the midpoint of $AC$, with a short computation, we have $PC \times PA = PR \times PM$, but $PC\times PA = PB^2 = PD^2 $, so $PR \times PM = PB^2 = PD^2$, hence $PB$ and $PD$ are the tangents in $B$ and $D$ to $(BMD)$. Therefore, $\angle BMR = \angle RMD = \angle PBD = \angle PDB$. But $\angle CDE = \angle CDB + \angle BDE = \angle CDB + \angle CBD =$ $\angle CDB + \angle CDP = \angle BDP = \angle BMC$, since $ED \parallel BC$ and $PD$ is tangent to $(ABCD)$ in $D$. So, $CMED$ is cyclic, hence $\angle CED = \angle CMD$, but $\angle CMD = \angle BDP = \angle BFD$, therefore $\angle BFE = \angle CED$, in other words, $CE \parallel BF$, as desired.

Its an easy complex bash, let's take the unit circle with $A,B,C \longrightarrow a,b,c$ and center $O \longrightarrow \vec{o}$ and observe that $ABCD$ is a harmonic quadrilateral so $D \longrightarrow d = \frac{2ac-ab-bc}{a+c-2b}$ and $F \longrightarrow f= \frac{bc}{d}=\frac{bc(a+c-2b)}{2ac-ab-bc}$ Now let $E' \longrightarrow e'= c+f-b$. Obviously, $E'$ lies on $DF$ and we claim that it does so on $BM$ as well. Indeed, $\frac{c+f-2b}{2b-a-c}=\frac{a(b-c)^2}{(2ac-ab-bc)(2b-a-c)}$ which on conjugating is invariant and so $B,M,E'$ are collinear. So $E' \equiv E$. $\square$

Sorry for any typos EMC 2014 Seniors P3 wrote: Let $ABCD$ be a cyclic quadrilateral in which internal angle bisectors $\angle ABC$ and $\angle ADC$ intersect on diagonal $AC$. Let $M$ be the midpoint of $AC$. Line parallel to $BC$ which passes through $D$ cuts $BM$ at $E$ and circle $ABCD$ in $F$ ($F \neq D$ ). Prove that $BCEF$ is parallelogram Solution: Obviously, $ABCD$ is harmonic. Now, Let $E' \in FD$, such $CE' || BF$. Let $X \in \odot (ABC)$ such, $BX||AC$ $\implies$ $X \in DM$ $\implies$ $\angle BXD$ $=$ $\angle CMD$ $=$ $\angle BFD$ $=$ $\angle CE'D$ $\implies$ $DCE'M$ is cyclic. Assume, $B,E',M$ position in the same order. Then, $\angle AMB$ $=$ $180^{\circ}-\angle BAD$ $=$ $\angle CBF$ $=$ $180^{\circ}$ $-$ $\angle CDE'$ $=$ $180^{\circ}$ $-$ $\angle CME'$ $\implies$ $E \equiv E'$ $\qquad \blacksquare$

gobathegreat wrote: Let $ABCD$ be a cyclic quadrilateral in which internal angle bisectors $\angle ABC$ and $\angle ADC$ intersect on diagonal $AC$. Let $M$ be the midpoint of $AC$. Line parallel to $BC$ which passes through $D$ cuts $BM$ at $E$ and circle $ABCD$ in $F$ ($F \neq D$ ). Prove that $BCEF$ is parallelogram Proposed by Steve Dinh Nice problem. The condition gives us $ABCD$ being a harmonic quadrilateral. Let $FA \cap BC = G$. Claim. $B$ is the midpoint of $GC$. Proof. $ F (A,C;B,D) = (\infty_{FD}, C; B, G) = -1$. Then $G$ must be the reflection of $C$ wrt $B$. We have $BG \equiv BC \parallel FD \equiv FE$. It suffices to prove that $FE = BC$ to prove that $BCFE$ is a parallelogram, which is equivalent to prove that $FE = BC = BG$. So, we need to prove that $BGEF$ is a parallelogram. But it suffices to prove that $BE \parallel FG$, which is true since $BM \parallel AG$, since $B$ is the midpoint of $CG$ and $M$ is the midpoint of $AC$. Done!

Nice one, I didn't use projective geo to solve it. I have a different and beautiful solution to this one. Notice that we just need to prove $BM || AF$, where $M$ is the midpoint of $AC$ (by midbase we're done). Now, redefine the problem as: Let $ABC$ be a triangle, $M$ the midpoint of $AC$, $AF || BM$ and $F \in (ABC)$, the parallel from $F$ wrt $BC$ touches $(ABC)$ in $D$. Prove $DA/DC=BA/BC$. Now define $H$ the point where $ABCH$ is an isosceles trapezoid. As $DC=FB, DA=FH$ we must prove $FB/FH=BC/BA$ and we get rid of $D$ :0. Now it is angle chase to prove $\angle FHB=\angle MBA, \angle FBH=\angle MBC$ and by law of sines on $MAB,MAC$ we're done.

Let $S_1$ be the midpoint of arc (ADC) and $S_2$ be the midpoint of arc (ABC). Furthermore let $S$ = $AC \cap BS_1 \cap DS_2$ (By Problem condition) Let $B'$ be the point on the circle such that DB' || AC Claim 1: $BMB'$ collinear We have \[ -1 = (AC;S_1S_2) \stackrel{S}{=} (AC;BD) \stackrel{B'}{=} (AC; AC \, \cap \, BB', P_{\infty}) \implies AC \, \cap \, BB' = M \] Claim 2: $S_1DSM$ are cyclic We have: \[ \measuredangle SMS_1 = 90 = \measuredangle S_2DS_1 = \measuredangle SDS_2 \] Claim 3: $CDME$ are cyclic We have: \[ \measuredangle EMC = \measuredangle BB'D = \measuredangle BFD = \measuredangle EDC \]Note that $MC$ bisects $\angle EMD$. Claim 4: $BF || CE$ Angle chase: \[ \measuredangle CED = \measuredangle CMD = \measuredangle BMC = \measuredangle BB'D = \measuredangle BFD \] Therefore we are done.