It can be proved by using vectors. Actually the same idea was used in IMO 2003 #3.

I had a similar idea, but I was very DUMB not to see what happened if I went on this path. I got the INEQUALITIES, but I thought that adding them would never give something useful: $\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$. (I don't know why, but this inequality seemed too natural for me to yield anything; really don't know why) Squaring and adding: $\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF} \right)^2 \leq 0$. Thus, $BC \| EF$ etc. Also, $BC \| DA$. The conclusion is now obvious. Thanks, Sung-yoon Kim.

Yeah I think you're on the right way.. Quote: $\left| \overrightarrow{AD} \right| = \left| \overrightarrow{BC} \right| + \left| \overrightarrow{EF} \right| \geq \left| \overrightarrow{FC} + \overrightarrow{BE} \right|$ Consider a parellelogram BCFX. Then $BC+EF=XF+FE \geq XE=|\overrightarrow{XB}+\overrightarrow{BE}|=|\overrightarrow{CF}+\overrightarrow{BE}|$. The rest is same as yours.

Guys I think we shoulld find sinthetical solution for this one,I havn't do it still ,but I will try again .

Yes it is a good idea to find a synthetic proof for this problem. Davron

That's really hard to find a such solution.

Yes,it is realy hard job,if that can help you I know that ther are sintetic solution,I know a man who has found it,but I want find it from misalf,so I havn't ask him about that solution.

perfect_radio wrote: The conclusion is now obvious. Well, after proving all these parallels, I still can't see why it's obvious...

Let $AD \cap CF = \left\{ T \right\}$. The following statements hold: - $ABCT$ is a parallelogram; - $DEFT$ is a parallelogram; - $\triangle ATF \sim \triangle DTC$; - the conclusion.

perfect_radio wrote: Also, $BC \| DA$. And why is this?

$\left( \overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}\right)^{2}\leq 0$ implies $\overrightarrow{AD}+\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow 0$. Also note that $\overrightarrow{EB}+\overrightarrow{CF}= \overrightarrow{CB}+\overrightarrow{EF}$. Hence, \[\overrightarrow{AD}= \overrightarrow{BC}+\overrightarrow{FE}.\]

A simple construction of the configuration, as the problem states, is as follows: A triangle $\bigtriangleup KLM$ is given and let $A$ be, a fixed point on the extension of the sideline $KL$ $($ $K,$ between $A,$ $L$ $).$ The line through the $A$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $B.$ The line through the $B$ and parallel to $KL,$ intersects the sideline $LM$ at a point, so be it $C.$ The line through the $C$ and parallel to $KM,$ intersects the sideline $KL$ at a point, so be it $D.$ The line through the $D$ and parallel to $LM,$ intersects the sideline $KM$ at a point, so be it $E.$ Through $A,$ $E$ now, we draw to lines parallel to $KM\equiv BE$ and $KL\equiv AD$ respectively, which intersect the sideline $LM,$ at points $F,$ $F'$ and we will prove that $F'\equiv F.$ It is easy to show that $CF = CM+MF = DE+AB$ $,(1)$ and $CF' = CL+LF' = AB+DE$ $,(2)$ From $(1),$ $(2)$ $\Longrightarrow$ $CF' = CF$ $\Longrightarrow$ $F'\equiv F.$ So, it has already been constructed the configuration as the problem states. $\bullet$ From $AB\parallel DE$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE}$ $,(3)$ But, $\frac{KA}{KD}= \frac{EF}{BC}$ $,(4)$ and $\frac{KB}{KE}= \frac{CD}{AF}$ $,(5)$ From $(3),)$ $(4),$ $(5)$ $\Longrightarrow$ $\frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC}$ and the proof is completed. Kostas Vittas.

Attachments:

Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote $a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$. Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$, from triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$. We will remember Euler theorem for quadrilateral and midpoint of opposite sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$. Equality is hold when $BC \parallel AD$ and our lemma is proved. Now we will solve the problem. Denote $AB=a,BC=b,CD=c, DE=d, EF=e, AF=f, AC=x_{1}, BD=x_{2}, CE=x_{3}, DF=x_{4}, EA=x_{5}, FB=x_{6}, AD=l_{1}, BE=l_{2}, CF=l_{3}$. We have $l_{1}=b+e, l_{2}= c+f, l_{3}=a+f$. Apply our theorem for $ABCD$ and $DEFA$ we have $a^{2}+c^{2}+2bl_{1}\geq x_{1}^{2}+x_{2}^{2}$ and $d^{2}+f^{2}+2el_{1}\geq x_{4}^{2}+x_{5}^{2}$. Adding all of them: \[a^{2}+c^{2}+d^{2}+f^{2}+2l_{1}^{2}\geq x_{1}^{2}+x_{2}^{2}+x_{4}^{2}+x_{5}^{2}\] Summing all such quadrilaterals we get \[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})\geq 2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2}) \] Next step is to apply our lemma for quadrilateral $BCEF$: $x_{3}^{2}+x_{6}^{2}+2be\geq l_{2}^{2}+l_{3}^{2}$. Summing with inequalities for $CDAF$, $DEAB$: \[(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+x_{5}^{2}+x_{6}^{2})+2ad+2be+2cf\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2}) \] Multiplying second inequality by $2$ and adding with first one we get \[2(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}+f^{2})+4(ad+be+cf)\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2}) \] or just $2(a+d)^{2}+2(b+e)^{2}+2(c+f)^{2}\geq 2(l_{1}^{2}+l_{2}^{2}+l_{3}^{2})$ As we see the equality is hold and thus from lemma's case of equality we get $BC \parallel AD \parallel EF$ and similarly other lines are parallel. Let a line through $C$ parallel to $AB$ intersects $AD$ at $P$, then $AP=b$ and thus $PD=e$ and $PD\parallel EF$. Hence $PDEF$ is parallelogram and $C,P,F$ are collinear as are $A,P,D$. So triangles $APF$ and $DPC$ are similar and from here \[\frac{AP}{DP}=\frac{PF}{PC}=\frac{AF}{CD}\] Rewriting in hexagon's sides: \[\frac{BC}{EF}=\frac{DE}{AB}=\frac{AF}{CD}\]

This was problem was proposed for Mathematical Reflections 2 in 2006, by Nairi Sedrakyan (Armenia). Mathematical Reflections: http://www.awesomemath.org

in a convex hexagon with AD=BC+EF, CF=DE+AB,BE=AF+CD.THERE WE CAN FIND EASILY.A simple construction of the configuration, as the problem states, is as follows: A triangle \bigtriangleup KLM is given and let A be, a fixed point on the extension of the sideline KL ( K, between A, L ). The line through the A and parallel to LM, intersects the sideline KM at a point, so be it B. The line through the B and parallel to KL, intersects the sideline LM at a point, so be it C. The line through the C and parallel to KM, intersects the sideline KL at a point, so be it D. The line through the D and parallel to LM, intersects the sideline KM at a point, so be it E. Through A, E now, we draw to lines parallel to KM\equiv BE and KL\equiv AD respectively, which intersect the sideline LM, at points F, F' and we will prove that F'\equiv F. It is easy to show that CF = CM+MF = DE+AB ,(1) and CF' = CL+LF' = AB+DE ,(2) From (1), (2) \Longrightarrow CF' = CF \Longrightarrow F'\equiv F. So, it has already been constructed the configuration as the problem states. \bullet From AB\parallel DE \Longrightarrow \frac{AB}{DE}= \frac{KA}{KD}= \frac{KB}{KE} ,(3) But, \frac{KA}{KD}= \frac{EF}{BC} ,(4) and \frac{KB}{KE}= \frac{CD}{AF} ,(5) From (3),) (4), (5) \Longrightarrow \frac{AB}{DE}= \frac{CD}{AF}= \frac{EF}{BC} and the proof is completed.

prowler wrote: Lemma: Let $ABCD$ be arbitrary quadrilateral. Denote $a,b,c,d$ it's sides and $e,f$ diagonals, then $a^{2}+c^{2}+2bd\geq e^{2}+f^{2}$. Proof: Denote $M, N, P$ midpoints of $BC, AD, BD$, from triangle inequality $MN\leq MP+NP$ or $2MN\leq b+d$. We will remember Euler theorem for quadrilateral and midpoint of opposite sides or diagonals: $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}\leq a^{2}+c^{2}+(b+d)^{2}$. Equality is hold when $BC \parallel AD$ and our lemma is proved. Sorry for reviving the old topic, but your lemma is not true, and therefore the whole solution as well. From Euler theorem we have not as you wrote $b^{2}+d^{2}+e^{2}+f^{2}=a^{2}+c^{2}+4MN^{2}$ but $ a^{2}+b^{2}+c^{2}+d^{2}=e^{2}+f^{2}+4MN^{2} $. So your lemma rewrites as $ e^2+f^2+2bd\geq a^2+c^2 $.