Giải thích các bước giải:
1.Ta có :
$NC//MD\to\dfrac{MF}{FC}=\dfrac{MD}{NC}=\dfrac{AN}{NC}=\dfrac{BD}{DC}=\dfrac{BM}{MA}=\dfrac{BM}{NA}$
Vì $ND//AB,MD//AC,ABCD$ là hình vuông $\to AN=ND=DM=MA$
Do $\dfrac{MF}{FC}=\dfrac{BM}{NA}=\dfrac{BM}{ND}=\dfrac{EM}{ED} $
$\to EF//BC$
2.Ta có :
$\dfrac{MF}{FC}=\dfrac{BM}{NA}\to\dfrac{BM}{MA}=\dfrac{FD}{FN}$
$\to\dfrac{BM+MA}{MA}=\dfrac{FD+FN}{FN}$
$\to\dfrac{BA}{MA}=\dfrac{ND}{FN}$
$\to \dfrac{BA}{AN}=\dfrac{NA}{NF}$
Mà $\widehat{ANF}=\widehat{BAN}=90^o\to\Delta NAF\sim\Delta ABN(c.g.c)$
$\to \widehat{FAN}=\widehat{ABN}\to AF\perp NB$
Tương tự $AE\perp MC$
Mà $MC\cap NB=K\to K$ là trực tâm $\Delta AEF\to AK\perp EF\to AK\perp BC(BC//EF)$
3.Ta có : $\Delta MDB\sim\Delta NCD(g.g)\to\dfrac{MD}{NC}=\dfrac{MB}{ND}=\dfrac{DB}{DC}$
$\to\dfrac{a}{NC}=\dfrac{MB}{a}$
$\to NC.MB=a^2$
$\begin{split}\to BC^2&=AC^2+AB^2\\&=(AN+NC)^2+(AM+MB)^2\\&= (a+NC)^2+(a+MB)^2\\&=2a^2+2a(NC+MB)+MB^2+NC^2\\&\ge 2a^2+2a.2\sqrt{NC.MB}+2MB.NC\\&=2a^2+2a.2a+2a^2\\&=8a^2\end{split}$
$\to BC\ge 2a\sqrt{2}$
Dấu = xảy ra khi $NC=MB\to AD\perp BC$