a,b.......
a, GS [TEX]\widehat{ABC}=\widehat{BCA}=\widehat{DME}=\alpha[/TEX]
Ta có :[TEX]\widehat{BMD}+\widehat{EMC}=180^o-\alpha[/TEX]
Lại có: [TEX]\widehat{BMD}+\widehat{BDM}=180^o-\alpha[/TEX] ( tổng 3 góc 1 t/giác)
[TEX]\Rightarrow \ \widehat{EMC}=\widehat{BDM}[/TEX]
Xét [TEX]\triangle \ \ BDM[/TEX] và [TEX]\triangle \ \ CME[/TEX] có :
[TEX]\widehat{EMC}=\widehat{BDM} \ ; \ \widehat{ABC}=\widehat{BCA}[/TEX]
[TEX]\Rightarrow \ \triangle \ \ BDM \ \sim \ \ \triangle \ \ CME[/TEX] (g-g)
[TEX]\Rightarrow \ \frac{BD}{CM}=\frac{BM}{CE}\Rightarrow \ BD.CE=a^2[/TEX]
Vậy BD.CE ko đổi.
b, [TEX]\frac{DM}{ME}=\frac{BD}{CM}=\frac{BD}{BM}[/TEX]
Xét [TEX]\triangle \ \ DEM[/TEX] và [TEX]\triangle \ \ DMB[/TEX] có :
[TEX]\widehat{ABC}=\widehat{DME} \ ; \ \frac{DM}{ME}=\frac{BD}{BM}[/TEX]
[TEX]\Rightarrow \ \triangle \ \ DEM \ \sim \ \ \triangle \ \ DMB[/TEX] (c-g-c)
[TEX]\Rightarrow \ \widehat{MDE}=\widehat{BDM}[/TEX] (2 góc t/ứ)
Vậy DM là phân giác [TEX]\widehat{BDE}[/TEX]