Giải thích các bước giải:

a.Xét $\Delta CHM,\Delta CDA$ có:

Chung $\hat C$

$\widehat{CHM}=\widehat{CDA}(=90^o)$

$\to \Delta CHM\sim\Delta CDA(g.g)$

$\to \dfrac{CH}{CD}=\dfrac{CM}{CA}$

$\to CM.CD=CH.CA$

b.Xét $\Delta CHB,\Delta CBA$ có:

Chung $\hat C$

$\widehat{CHB}=\widehat{CBA}(=90^o)$

$\to \Delta CHB\sim\Delta CBA(g.g)$

$\to \dfrac{CH}{CB}=\dfrac{CB}{CA}$

$\to CB^2=CH.CA$

$\to BC^2=CM.CD$

c.Xét $\Delta CMB,\Delta CBD$ có:
Chung $\hat C$

$\dfrac{BC}{CM}=\dfrac{CD}{CB}$ vì $CB^2=CM.CD$

$\to \Delta CMB\sim\Delta CBD(c.g.c)$

$\to \widehat{CDB}=\widehat{CBM}=90^o-\widehat{BMC}=90^o-\widehat{DMF}=\widehat{MDF}$

$\to DC$ là phân giác $\widehat{BDF}$

d.Vì $ DC$ là phân giác $\widehat{BDF}$

$\to DM$ là phân giác $\widehat{BDF}$

$\to \dfrac{MF}{MB}=\dfrac{DF}{DB}$

Ta có:

$\sin\widehat{DBF}=\dfrac{DF}{DB}$

$\to \sin\widehat{DBF}=\dfrac{FM}{MB}$