Question

[Thị Hà Lò]

Giúp nhanh ạ, e cảm ơn

Answer 1

[Trần Tuấn Hoàng]

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 \(\widehat{AMB}=120^0\)