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+ Ta có $\widehat{C_1} + \widehat{C_2} = \widehat{ABC} $. Hay $ \widehat{C_1} + \widehat{C_2} = 60^o \Longrightarrow \widehat{C_1} = 60^o- \widehat{C_2} $
+ Ta có $\widehat{C_3} + \widehat{C_2} = \widehat{MCD} $. Hay $ \widehat{C_3} + \widehat{C_2} = 60^o \Longrightarrow \widehat{C_2} = 60^o- \widehat{C_3} $
+ Xét $\Delta ACM$ và $\Delta BCD$ có
AC=BC ($\Delta ABC$ đều)
CM=DC ($\Delta MDC$ đều)
$\widehat{C_2} = \widehat{C_1} =60^o- \widehat{C_3} $
$\Longrightarrow \Delta ACM= \Delta BCD$ (cgc) (đpcm)
b, + $\Delta MDC đều \Longrightarrow MC=MD = 2$
+ Theo phần a: $\Delta ACM= \Delta BCD \Longrightarrow AM=BD= 1$
+ Xét $\triangle MBD$ ta có $MB^2+BD^2= (\sqrt{3})^2+1^2= 4 = 2^2= MD^2$
$\Longrightarrow \Delta MDB$ vuông ở B (theo pytago đảo)
c, + $\Delta MDB$ vuông ở B có $BD= \dfrac{MD}{2}$ ~~> $ \widehat{M_3}=30^o; \widehat{D_1} = 60^o $
+ Ta có $ \widehat{BMC}= \widehat{M_3}+ \widehat{M_2}= 30^o+60^o= 90^o$
+ Ta có $ \widehat{BDC}= \widehat{D_1}+ \widehat{D_2}= 60^o+60^o= 120^o$
Mà $ \widehat{BDC}= \widehat{M_4}= 120^o$ ($ \Delta ACM= \Delta BCD$)
+ Ta có $ \widehat{BMC}+ \widehat{M_4}+ \widehat{M_1}=360^o$. Hay $ 90^o+ 120^o+ \widehat{M_1}=360^o \Longrightarrow \widehat{M_1}=150^o $
+ Ta có $\widehat{M_1}+ \widehat{M_3}= 150^o+30^o= 180^o =\widehat{AMD}$
$ \Longrightarrow 3$ điểm A;M;D thẳng hàng (đpcm)
d, + Theo phần c ta có $ \widehat{BMC}=90^o$ nên $\Delta BMC$ vuông ở M nên
$MB^2+MC^2=BC^2 = (\sqrt{3})^2+2^2= 3+4=7$
$ \Longrightarrow BC=\sqrt{7} $