BM=CF and <MBC=<FCA means ABC is equilateral

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WakeUp

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WakeUp

#1 h Oct 1, 2011, 3:08 PM • 1 Y

Y by Adventure10

In an acute-angled triangle $ABC$ , $CF$ is an altitude, with $F$ on $AB$ , and $BM$ is a median, with $M$ on $CA$ . Given that $BM=CF$ and $\angle MBC=\angle FCA$ , prove that triangle $ABC$ is equilateral.

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goodar2006

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goodar2006

#2 h Oct 1, 2011, 3:49 PM • 2 Y

Y by Adventure10, Mango247

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.

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MariusBocanu

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MariusBocanu

#3 h Oct 1, 2011, 4:04 PM • 2 Y

Y by Adventure10, Mango247

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.

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goodar2006

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goodar2006

#4 h Oct 1, 2011, 4:16 PM • 2 Y

Y by Adventure10, Mango247

MariusBocanu wrote:

....... $CT=TB$ ,.....

why??

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MariusBocanu

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MariusBocanu

#5 h Oct 1, 2011, 4:21 PM • 2 Y

Y by Adventure10, Mango247

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

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vanu1996

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vanu1996

#6 h Aug 17, 2013, 5:06 PM • 3 Y

Y by adam.romanov, Adventure10, Mango247

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

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adam.romanov

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adam.romanov

#7 h Jan 2, 2016, 11:35 PM • 2 Y

Y by Adventure10, Mango247

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

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goodar2006

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goodar2006

#8 h Jan 4, 2016, 5:13 AM • 1 Y

Y by Adventure10

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.

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dgmath97

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dgmath97

#9 h Jan 4, 2016, 1:28 PM • 2 Y

Y by Adventure10, Mango247

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.....

Attachments:

This post has been edited 8 times. Last edited by dgmath97, Jan 4, 2016, 1:42 PM

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Mmesomachi

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Mmesomachi

#10 h Jan 25, 2017, 10:57 PM • 2 Y

Y by Adventure10, Mango247

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

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Tsikaloudakis

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Tsikaloudakis

#11 h Jun 18, 2022, 2:33 PM

Y by

See the figure:

Attachments:

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