Giải thích các bước giải:

a.Ta có : $DM\perp BC\to\widehat{CMD}=\widehat{CAB}=90^o$ 

$\to \Delta ABC\sim\Delta MDC(g.g)$

b.Ta có : $\widehat{BAC}=\widehat{BMI}=90^o$

$\to\Delta BMI\sim\Delta BAC(g.g)$

$\to\dfrac{BM}{BA}=\dfrac{BI}{BC}\to BI.BA=BM.BC$

c.Ta có : $BA.BI=BM.BC\to \dfrac{BA}{BC}=\dfrac{BM}{BI}$

$\to \Delta BAM\sim\Delta BCI(c.g.c)$

$\to \widehat{BAM}=\widehat{BCI}$

Tương tự chứng minh được $\widehat{IAK}=\widehat{ICB}$

$\to \widehat{IAK}=\widehat{BAM}$

$\to 90^o-\widehat{IAK}=90^o-\widehat{BAM}$

$\to \widehat{KAC}=\widehat{CAM}$

$\to AC$ là phân giác $\widehat{KAM}$

d.Ta có : $BC^2=AB^2+AC^2=100\to BC=10$

Từ câu a $\to \dfrac{CB}{CD}=\dfrac{CA}{CM}=\dfrac{AB}{MD}$

$\to CM.CB=CD.CA, MD=\dfrac{AB.CD}{CB}$

$\to CD=\dfrac{CM.CB}{CA},MD=\dfrac{AB.CM}{CA}$

$\to AD=AC-CD=AC-\dfrac{CM.CB}{CA}$

$\to S_{ABMD}=S_{ABD}+S_{BMD}$

$\to S_{ABMD}=\dfrac12AB.AD+\dfrac12BM.DM$

$\to S_{ABMD}=\dfrac12AB(AC-\dfrac{CM.CB}{CA})+\dfrac12(BC-CM).\dfrac{AB.CM}{CA}$

$\to S_{ABMD}=\dfrac12\cdot 8(6-\dfrac{CM.10}{6})+\dfrac12(10-CM).\dfrac{8.CM}{6}$

$\to S_{ABMD}=-\dfrac{2CM^2}{3}+24$