\Rightarrow\Rightarrow\Rightarrow\Rightarrow
a) Ta có: [TEX]\widehat{AMC}= 60^o[/TEX]
[TEX]\widehat{AMC}= 60^o[/TEX]
\Rightarrow [TEX]\widehat{CMD}= 180^o - (60^o.2) = 60^o[/TEX]
\Rightarrow [TEX]\widehat{AMD}= 60^o + 60^o = 120o[/TEX]
\Rightarrow [TEX]\widehat{BMC}= 60^o + 60^o = 120o[/TEX]
Xét [TEX]\triangle AMD[/TEX] và [TEX]\triangle BMC[/TEX] có:
AM = CM
[TEX]\widehat{AMD} = \widehat{BMC}[/TEX]
MD = MB
Do đó: [TEX]\triangle AMD = \triangle BMC[/TEX]
b) Ta co [TEX]\triangle AMD=\triangle CMB[/TEX] ( kq cau a)
\Rightarrow [TEX]\widehat{FBM}=\widehat{EDM}[/TEX] ( 2 goc tuong ung)
va AD=CB ( 2 canh tuong ung) \Rightarrow AE:2=CB:2
lai co E, F lan luot la trung diem cua AD, BC (gt)
\Rightarrow AE=ED=FC=FB
Xet [TEX]\triangle EMD[/TEX] va[TEX]\triangle FMB][/TEX]
co ED=FB ( c/m tren)
[TEX]\widehat{EDM}=\widehat{FBM}[/TEX] (c/m tren)
MD=MB ([TEX]\triangle BMD[/TEX] deu)
nen [TEX]\triangle EMD=\triangle FMB[/TEX] (c.g.c)
\Rightarrow EM = FM ( 2 canh tuong ung)
do do [TEX]\triangle MEF[/TEX] can tai M ( dinh nghia tam giac can)
lai co [TEX]\widehat{EMD}=\widehat{FMB}[/TEX] ( 2 goc tuong tung cua tg EMD=tg FBM)
[TEX]\widehat{FMB}+\widehat{DMF}=60^o[/TEX] ( tg BMD deu)
\Rightarrow [TEX]\widehat{EMD}+\widehat{DMF}=60^o[/TEX]
hay [TEX]\widehat{EMF}=60^o[/TEX]
trong tam giac MEF can tai M co [TEX]\widehat{EMF}=60^o[/TEX]
Vay tam goac MEF la tam giac deu dpcm