equal angles, starting with an equailateral triangle

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equal angles, starting with an equailateral triangle

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Source: Irmo 2018 p2 q8

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#1 h Sep 16, 2018, 9:37 AM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the midpoint of side $BC$ of an equilateral triangle $ABC$ . The point $D$ is on $CA$ extended such that $A$ is between $D$ and $C$ . The point $E$ is on $AB$ extended such that $B$ is between $A$ and $E$ , and $|MD| = |ME|$ . The point $F$ is the intersection of $MD$ and $AB$ . Prove that $\angle BFM = \angle BME$ .

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#2 h Sep 16, 2018, 11:48 AM • 3 Y

Y by Maths_Guy, Adventure10, Mango247

It can be noted that $ME \neq MD$ for $AE = AD$ .

We have, $AM$ is the external bisector of $\angle EAD$ and $ME = MD$ . Thus $AMED$ is cyclic. Thus we have $\angle BFM = 180^{\circ} - \angle FME - \angle MEF = 180^{\circ} - \angle DAE - \angle MEF = 60^{\circ} - \angle MEF = \angle BME.$

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4381 posts

#3 h Sep 16, 2018, 4:21 PM • 3 Y

Y by Adventure10, Mango247, Mango247

Let $N,P$ the projections of $M$ onto $CA, AB$ respectively; as $MN=MP$ , with $MD=ME$ , we get $\triangle DMN\cong\triangle EMP$ , which implies $\angle ADM=\angle AEM \ (\ 1\ )$ , making $MADE$ cyclic, i.e. $\triangle ADF\sim\triangle MEF$ , done.

Best regards,
sunken rock

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819 posts

#4 h May 12, 2021, 8:11 AM • 2 Y

Y by Mango247, Mango247

Very easy by theorem 9 ( $Mr.vakili$ )

This post has been edited 1 time. Last edited by Hopeooooo, May 12, 2021, 8:11 AM
Reason: 2

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#5 h Aug 4, 2023, 3:25 PM

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Let angle BFM =x
angle DMC=60°+x, angle ACM =60°
angle MDC=60°-x
let angle BME=y, angle MBE=120°
angle BEM=60°-y
In ∆DMC,
MD/ MC=sin 60°/ sin(60°-x)
In ∆BME,
ME/ MB=sin 60°/ sin(60°-y)
Since, MB=MC
MD=ME
sin 60°/ sin(60°-x)=sin 60°/ sin(60°-y)
sin(60°-x) =sin(60°-y)
60°-x=60°-y
x=y
So, angle BFM=angle BME
@KRISHIJIVI

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#6 h Aug 30, 2023, 12:44 PM

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Let $G$ be a point on line $EG$ such that line $EG$ is parallel to line $BC$ and $MG=ME=MD$
⟹ $M$ is the center of a circle with point $E,D,G$ on it.
extend ray $EM$ to meet the circle at point $H$ (Not $E$ )
We have to prove $\angle BFM = \angle BME$
but $\angle BME = \angle MEG =\angle HEG =\angle HDG$
also $\angle MGD =\angle MDG =60-\angle MEG$
$\therefore \angle MDH = 60$
$\therefore$ since $\angle EDH =90, \angle EDM =30$
$\therefore \angle DME = 120$
$\therefore$ since $\angle MEF = 60-\angle MEG, \angle EFM= MEG, \therefore \angle BFM =\angle BME$
$Q.E.D$
I also thought this could be solved via barycentric coordinates but couldn't figure it out exactly

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#8 h Apr 7, 2025, 6:32 PM

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see the figure

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