$1)$

Xét $\triangle ABC$ vuông tại $A$ có:

$+)$ $BC^2 = AB^2 + AC^2$ $($Đlý Pythagore$)$

$BC^2 = 6^2 + 8^2$

$BC^2 = 36 + 64$

$BC^2 = 100$

$BC = \sqrt{100} = 10 \text{ cm}$

$+)$  $\sin C = \dfrac{AB}{BC} \quad \text{(TSLG)}$ 

$\sin C = \dfrac{6}{10} = 0,6$

$\Rightarrow \widehat{C} \approx 36^\circ 52'$

$+)$  $\sin B = \dfrac{AC}{BC} \quad \text{(TSLG)}$

$\sin B = \dfrac{8}{10} = 0,8$

$\Rightarrow \widehat{B} \approx 53^\circ 8'$

Vậy ....

 

$2)$

$a)$

Ta có:

+) $BD \perp AM$ tại $D \Rightarrow \widehat{BDA} = 90^\circ$

+) $CE \perp AM$ tại $E \Rightarrow \widehat{AEC} = 90^\circ$

Xét $\triangle BAD$ và $\triangle ACE$ có:

+) $\widehat{BDA} = \widehat{AEC} = 90^\circ$ $(\text{cmt})$

+) $\widehat{ABD} = \widehat{CAE}$ $($cùng phụ với $\widehat{BAD}$$)$

$\Rightarrow \triangle BAD \sim \triangle ACE$ $(g.g)$

  

$b)$

Xét $\triangle ABC$ vuông tại $A$ có:

$\tan \widehat{ACB} = \dfrac{AB}{AC} \quad \text{(TSLG)}$

Vì $\triangle BAD \sim \triangle ACE$ ($\text{cmt}$)

$\Rightarrow \dfrac{BD}{AE} = \dfrac{AB}{AC}$

$\Rightarrow \dfrac{BD}{AE} = \tan \widehat{ACB}$

$\Rightarrow BD = AE \cdot \tan \widehat{ACB} \quad (1)$ (đpcm)

Vì $\triangle BAD \sim \triangle ACE$ ($\text{cmt}$)

$\Rightarrow \dfrac{AD}{CE} = \dfrac{AB}{AC}$

$\Rightarrow \dfrac{AD}{CE} = \tan \widehat{ACB}$

$\Rightarrow CE = \dfrac{AD}{\tan \widehat{ACB}} \quad (2)$

Vì $\begin{cases} BD \perp AM \\ CE \perp AM \end{cases}$

$\Rightarrow BD \parallel CE$ 

$\Rightarrow \widehat{MBD} = \widehat{MCE}$ $(2$ góc ở vị trí so le trong$)$

Xét $\triangle MBD$ và $\triangle MCE$ có:

+) $\widehat{BDM} = \widehat{CEM} = 90^\circ \text{ (cmt)}$

+) $\widehat{MBD} = \widehat{MCE}$ $(\text{cmt})$

$\Rightarrow \triangle MBD \sim \triangle MCE \text{ (g - g)}$

$\Rightarrow \dfrac{MB}{MC} = \dfrac{BD}{CE} \quad (3)$

Thay $(1), (2)$ vào $(3)$

$\Rightarrow \dfrac{MB}{MC} = \dfrac{AE \cdot \tan \widehat{ACB}}{\dfrac{AD}{\tan \widehat{ACB}}}$

$\Rightarrow \dfrac{MB}{MC} = \dfrac{AE \cdot \tan \widehat{ACB} \cdot \tan \widehat{ACB}}{AD}$

$\Rightarrow \dfrac{MB}{MC} = \dfrac{AE \cdot \tan^2 \widehat{ACB}}{AD}$ (đpcm)

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