1) DC cắt AB tại H.
- Ta có: \(\widehat{DAB}+\widehat{BAC}=\widehat{DAC}\) ; \(\widehat{CAE}+\widehat{BAC}=\widehat{BAE}\).
Mà \(\widehat{DAB}=\widehat{CAE}=60^0\) (△ABD đều, △ACE đều).
=>\(\widehat{DAC}=\widehat{BAE}\)
- Xét △DAC và △BAE có:
\(\left[{}\begin{matrix}AD=AB\left(\Delta ABDđều\right)\\\widehat{DAC}=\widehat{BAE}\left(cmt\right)\\AC=AE\left(\Delta ACEđều\right)\end{matrix}\right.\)
=>△DAC = △BAE (c-g-c).
=>\(\widehat{ADC}=\widehat{ABE}\) (2 góc tương ứng).
- Ta có: \(\widehat{ADH}+\widehat{HAD}+\widehat{AHD}=180^0\) (tổng 3 góc trong △DAH).
\(\widehat{MBH}+\widehat{BMH}+\widehat{BHM}=180^0\) (tổng 3 góc trong △BMH).
Mà \(\widehat{ADH}=\widehat{MBH}\) (cmt) ; \(\widehat{BHM}=\widehat{AHD}\) (đối đỉnh).
=>\(\widehat{DAH}=\widehat{HMB}\) mà \(\widehat{DAH}=60^0\) (△ABD đều).
=>\(\widehat{HMB}=60^0\).
Mà \(\widehat{HMB}+\widehat{BMC}=180^0\) (kề bù).
=>\(60^0+\widehat{BMC}=180^0\)
=>\(\widehat{BMC}=120^0\).
2) Ta có: MF=MB (gt) nên △MBF cân tại M.
Mà \(\widehat{FMB}=60^0\) (cmt) nên △MBF đều.
=> \(\widehat{FBM}=60^0\) mà \(\widehat{ABD}=60^0\) (△ABD đều) nên \(\widehat{FBM}=\widehat{ABD}=60^0\)
Mà \(\widehat{FBH}+\widehat{ABM}=\widehat{FBM}\); \(\widehat{FBH}+\widehat{DBF}=\widehat{ABD}\).
=>\(\widehat{DBF}=\widehat{ABM}\)
- Xét △BFD và △BMA có:
\(\left[{}\begin{matrix}BD=BA\left(\Delta ABDđều\right)\\\widehat{DBF}=\widehat{ABM}\left(cmt\right)\\BF=BM\left(\Delta BMFđều\right)\end{matrix}\right.\)
=>△BFD = △BMA (c-g-c).
3) - Ta có: \(\widehat{DFB}+\widehat{BFM}=180^0\) (kề bù).
Mà \(\widehat{BFM}=60^0\) (△BFM đều) nên \(\widehat{DFB}+60^0=180^0\)
=>\(\widehat{DFB}=120^0\) mà \(\widehat{DFB}=\widehat{AMB}\) (△BFD = △BMA)
Nên \(\widehat{AMB}=120^0\)