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a) Ta có: [tex]\widehat{ABM}=\widehat{ACB}(=\widehat{BAx})[/tex]; góc $A$ chung
[tex]\Rightarrow \Delta ABM\sim \Delta ACB(g.g)\Rightarrow \frac{AB}{AM}=\frac{AC}{AB}\Rightarrow AB^2=AC.AM[/tex]
b) [tex]\widehat{ABM}=\widehat{ACB}\Rightarrow dpcm[/tex]