let $D$ be the feet of perpendicular from $B$ on $AC$. so we have $\angle ABD=\angle FCA=\angle MBC$, so the circumcenter $O$ lies on the line $BM$. since the triangle is acute, we must have that $BM$ is perpendicular to $AC$, so $AB=BC$. now we have $2S_{ABC}=BM.AC=AB.CF$ and since $BM=CF$, we get the desired result.

Or, note $T=BM\cap CF$, $CT=TB$, so $MT=TF$, so $MT.TB=CT.TF$, so $BCMF$ is concylic, so $\widehat{BMC}=\widehat{BFC}=90$, so $BC=BA$, but we also have $\widehat{BMF}=\widehat{BCF}=\widehat{MBC}$, so $MF\parallel BC$, so $F$ is midpoint of $AB$, so $CA=CB=AB$, so the triangle is equilateral.

MariusBocanu wrote: ....... $CT=TB$,..... why??

Sorry, i missread the question, i read $\widehat{MBC}=\widehat{FCB}$. Sorry again.

Join $MF$,so $MF=MC$,so ${\angle}MCF={\angle}MFC={\angle}MBC$,so $MFBC$ is cyclic,from here its easy to find the result.

goodar2006 wrote: let $D$ be the feet of perpendicular from $B$ on $AC$. so we have $\angle ABD=\angle FCA=\angle MBC$, so the circumcenter $O$ lies on the line $BM$. since the triangle is acute, we must have that $BM$ is perpendicular to $AC$, so $AB=BC$. now we have $2S_{ABC}=BM.AC=AB.CF$ and since $BM=CF$, we get the desired result. Could you please explain why : $\angle ABD=\angle FCA=\angle MBC$ leads to the conclusion circumcenter $O$ lies on the line $BM$ ? thanks

Try proving this, in a triangle $ABC$, if $H$ is the foot of the altitude from $A$ on $BC$, and $O$ the circumcenter of the triangle, then $\angle BAH = \angle CAO$. The proof is a simple angle-chasing. After proving it, conclude that the reverse is also true, which yields the conclusion I've drawn in my previous post.

My solution... The first part of the solution is in the below part of the attachment and the second part is in the upper part of the attachment...please forgive me for this.....

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Note FM is the median from F to AC in ΔFAC. Also, ΔFAC is right angled Therefore FM=AM=CM Also, because FM=MC ; <MFC=<MCF Also, <MFC=<MBC, hence FMCB is cyclic. This implies <MCF=<MBF Hence, median BM is also an altitude. Therefore AB=BC & <BAC=<BCA But chord FC=BM in cyclic quadrilateral FMCB, therefore <FBC=<BCM; which in turn implies that <ABC=<BCA now we have <ABC=<BCA=<BAC. Hence, ΔABC is indeed equilateral. Q.E.D

See the figure:

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