Giải thích các bước giải:
a.Ta có : $DM\perp AB, DN\perp AC\to \widehat{BMD}=\widehat{DNC}=90^o, DM//AC, DN//AB$
$\to \widehat{MDB}=\widehat{NCD}\to \Delta BMD\sim\Delta DNC(g.g)$
$\to \dfrac{BD}{DC}=\dfrac{MB}{ND}\to BD.ND=BM.CD$
b.Ta có : $AB\perp AC, DN\perp AC, DM\perp AB, AD$ là phân giác góc A $\to AMDN$ là hình vuông$\to DN=AN=AM=DM$
Ta có : $DM//AC\to BE//CN\to \dfrac{DE}{CN}=\dfrac{BD}{BC}$
$DN//AB\to DF//BM\to\dfrac{DF}{BM}=\dfrac{CD}{BC}$
$\to \dfrac{DE}{CN}:\dfrac{DF}{BM}=\dfrac{BD}{BC}:\dfrac{CD}{BC}$
$\to \dfrac{DE}{CN}.\dfrac{BM}{DF}=\dfrac{BD}{CD}$
$\to \dfrac{DE}{DF}.\dfrac{BM}{CN}=\dfrac{BD}{CD}$
$\to \dfrac{DE}{DF}=\dfrac{BD}{CD}.\dfrac{CN}{BM}$
Từ câu a $\to \dfrac{CN}{DM}=\dfrac{CD}{BD}=\dfrac{DN}{BM}$
$\to \dfrac{CN}{DM}.\dfrac{DN}{BM}=\dfrac{CD^2}{BD^2}$
$\to \dfrac{CN}{BM}=\dfrac{CD^2}{BD^2} (DN=DM)$
$\to \dfrac{CN}{BM}.\dfrac{BD}{CD}=\dfrac{CD}{BD}$
Mà $\dfrac{CD}{BD}=\dfrac{AN}{CN}=\dfrac{DN}{CN}$
$\to \dfrac{DE}{DF}=\dfrac{CN}{DN}$
$\to \Delta DEF\sim\Delta NCD(c.g.c)$
$\to \widehat{EFD}=\widehat{NDC}\to EF//CD$
c.Ta có : $DN//AB\to\dfrac{NF}{AM}=\dfrac{CN}{CA}=\dfrac{DN}{AB}$
$\to \dfrac{NF}{AM}=\dfrac{DN}{AB}=\dfrac{AN}{AB}, AN=DN$
Mà $\widehat{BAN}=\widehat{ANF}=90^o$
$\to \Delta ABN\sim\Delta NAF(c.g.c)\to \widehat{FAN}=\widehat{ABN}\to AF\perp BN$
