Toán 7 Tam giác

a)
AB=AD(1)
AC=AE(2)
[tex]\widehat{BAD}=\widehat{CAE}\Leftrightarrow \widehat{BAD}+\widehat{BAC}=\widehat{CAE}+\widehat{BAC}\Leftrightarrow \widehat{CAD}=\widehat{BAE}[/tex] (3)
Từ (1), (2), (3) => ∆ABE = ∆ADC (ĐPCM)
b) suy ra từ câu a
[tex]\widehat{ADC}=\widehat{ABE}\Rightarrow \widehat{ADB}+\widehat{DBA}=180^o-\widehat{DAB}=(\widehat{ADB}-\widehat{ADB})+(\widehat{DBA}+\widehat{ABE})=180^o-\widehat{DMB}[/tex]
[tex]\Rightarrow \widehat{DMB}=\widehat{DAB}=60^o[/tex]
[tex]\Rightarrow \widehat{BMC}=180^o-\widehat{DAB}=120^o[/tex] (ĐPCM)