a)
Xét $\Delta ABM$ và $\Delta HCM$:
$\widehat{BAM}=\widehat{CHM}\,(=90^o)$
$\widehat{AMB}=\widehat{HMC}$ (đối đỉnh)
$\to \Delta ABM=\Delta HCM$ (g.g)
b)
$\Delta ABM=\Delta HCM$ (cmt)
$\to \widehat{ABM}=\widehat{HCM}$
mà $\widehat{ABM}=\widehat{MBC}$ (gt)
$\to \widehat{MBC}=\widehat{HCM}$ (đpcm)
c)
BM là phân giác của $\widehat{ABC}$ (gt)
$\to \dfrac{AM}{MC}=\dfrac{AB}{BC}$
Theo định lí Pytago: $BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10cm$
$\to \dfrac{AM}{MC}=\dfrac{6}{10}=\dfrac{3}{5}\to AM=\dfrac{3MC}{5}$
Ta có: $AM+MC=AC=8cm \to \dfrac{3MC}{5}+MC=8cm \to MC=5cm$
$\to AM=\dfrac{3MC}{5}=3cm$
