Câu a)
Trong $\triangle ABC$ có $\widehat{A}+\widehat{B}+\widehat{C}=180^0$
$\Rightarrow \frac{1}2\widehat{A}+\frac{1}2\widehat{B}+\frac{1}2\widehat{C}=90^0$
Trong $\triangle CMI$ có $\widehat{CMI}=90^0-\frac{1}2\widehat{C}=\frac{1}2\widehat{A}+\frac{1}2\widehat{B}$
$\Rightarrow \widehat{AMI}=180^0-\widehat{CMI}=180^0-\frac{1}2\widehat{A}-\frac{1}2\widehat{B}$
Trong $\triangle ABI$ có $\widehat{AIB}=180^0-\frac{1}2\widehat{A}-\frac{1}2\widehat{B}$
$\Rightarrow \widehat{AMI}=\widehat{AIB}$
Mà $\widehat{MAI}=\widehat{BAI}=\frac{1}2\widehat{A}$
$\Rightarrow \triangle AMI \sim \triangle AIB (g.g) (đpcm)$