a,+$\triangle ABC$ cân ở A nên: $ \widehat{ABC}=\widehat{BCA}=\frac{180^o- \widehat{BAC}}{2}=\frac{180^o- 100^o}{2}=40^o$
+$\triangle EBM$ đều $\Longrightarrow \widehat{EBM}=60^o $
+Ta có: $\widehat{EBC}=\widehat{EBM}+\widehat{MBC}=60^o+10^o=70^o$
+Ta có: $\widehat{EBC}=\widehat{EBA}+\widehat{ABC}$. hay: $70^o=\widehat{EBA}+40^o \Longrightarrow \widehat{EBA} =30^o$
+$\triangle EBC$ có: $\widehat{EBC}+\widehat{ECB}+\widehat{BEC}=180^o$.
Hay: $70^o+40^o+\widehat{BEC}=180^o \Longrightarrow \widehat{BEC}=70^o =\widehat{EBC}$
$\Longrightarrow \triangle CBE$ cân ở C
+$\triangle EAB$ có: $\widehat{EBA}+\widehat{BEA}+\widehat{EAB}=180^o$.
Hay: $30^o+70^o+\widehat{BAE}=180^o \Longrightarrow \widehat{BAE}=80^o$
+ Ta có: $\widehat{EAC}=\widehat{EAB}+\widehat{BAC}=80^o+100^o=180^o$
$\Longrightarrow $ 3 điểm E; A; C thằng hàng
b, + Ta có: $\widehat{ABC}=\widehat{B_2}+\widehat{B_3}$. Hay: $40^o=10^o+\widehat{B_2} \Longrightarrow \widehat{B_2} =30^o$
+$\widehat{EBM}=\widehat{B_2}+\widehat{B_1}$. Hay: $60^o=30^o+\widehat{B_1} \Longrightarrow \widehat{B_1} =30^o=\widehat{B_2}$
+Xét $\triangle EAB$ và $ \triangle MAB$ ta có:
$BM=EB$ ($\triangle EBM$ đều )
$\widehat{B_2}=\widehat{B_1}$ (CM trên)
$BA$ chung
$ \Longrightarrow$ $ \triangle EBA = \triangle MBA $ (cgc)
$ \Longrightarrow \widehat{BEA} = \widehat{BMA}=70^o $
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