Giải thích các bước giải:

a.Xét $\Delta ABE, \Delta ACF$ có:

Chung $\hat A$

$\hat E=\hat F(=90^o)$

$\to \Delta AEB\sim\Delta AFC(g.g)$

$\to \dfrac{AE}{AF}=\dfrac{AB}{AC}$

$\to \Delta AEF\sim\Delta ABC(c.g.c)$

b.Vì $BE\perp AC, CF\perp AB, BE\cap CF=H$

$\to H$ là trực tâm $\Delta ABC$

$\to AH\perp BC$

Mà $BM//CF\to BM\perp AB$

$\to \widehat{MBD}=90^o-\widehat{DBA}=\widehat{DAB}$

Mà $\widehat{BDM}=\widehat{BDA}(=90^o)$

$\to \Delta BDM\sim\Delta ADB(g.g)$

$\to \dfrac{BD}{DA}=\dfrac{DM}{DB}$

$\to DB^2=AD.DM$

c.Ta có: $\widehat{AHE}=90^o-\widehat{HAE}=90^o-\widehat{DAC}=\hat C=45^o$

$\to \Delta AEH$ vuông cân tại $E$

$\to \dfrac{AE}{AH}=\dfrac1{\sqrt2}$

Từ a $\to \widehat{AEK}=\widehat{AEF}=\widehat{ABC}=\widehat{ABD}=90^o-\widehat{DAB}=90^o-\widehat{HAF}=\widehat{AHF}$

Mà $\widehat{AKE}=\widehat{AFH}(=90^o)$

$\to \Delta AKE\sim\Delta AFH(g.g)$

$\to \dfrac{S_{AFH}}{S_{AKE}}=\dfrac{AE^2}{AH^2}=\dfrac12$