Giải thích các bước giải:

Bài 5:

a.Do $M$ là trung điểm $BC\to MB=MC=\dfrac12BC$

Vì $DF//AM$

$\to \dfrac{DE}{AM}=\dfrac{BD}{BM}=\dfrac{BD}{\dfrac12BC}=\dfrac{2BD}{BC}$

     $\dfrac{DF}{AM}=\dfrac{DC}{MC}=\dfrac{DC}{\dfrac12BC}=\dfrac{2DC}{BC}$

$\to \dfrac{DE}{AM}+\dfrac{DF}{AM}=2\cdot (\dfrac{DB}{BC}+\dfrac{DC}{BC})$

$\to \dfrac{DE+DF}{AM}=2$

$\to DE+DF=2AM$

b.Ta có: $AN//BC, AM//DF$

$\to AMDN$ là hình bình hành
$\to DN=AM$

Mà $DE+DF=2AM$

$\to DE+DF=2DN$

$\to DF-DN=DN-DE$

$\to FN=NE$

$\to N$ là trung điểm $EF$

c.Vì $DF//AM, AN//BC$

$\to \Delta FDC\sim\Delta AMC(g.g)$

     $\Delta FNA\sim\Delta FDC(g.g)$

$\to \dfrac{S_{FDC}}{S_{AMC}}=\dfrac{FD^2}{MA^2}$

     $\dfrac{S_{FDC}}{S_{FNA}}=\dfrac{FD^2}{FN^2}$

Ta có:

$DF^2=(FN+ND)^2\ge 4DN\cdot FN$

$\to DF^2\ge 4DN\cdot FN$

$\to DF^2\ge 4AM\cdot FN$

$\to \dfrac{DF}{AM}\cdot \dfrac{DF}{FN}\ge 4$

$\to \dfrac{DF^2}{AM^2}\cdot \dfrac{DF^2}{FN^2}\ge 16$

$\to \dfrac{S_{FCD}}{S_{AMC}}\cdot \dfrac{S_{FDC}}{S_{FNA}}\ge 16$

$\to S^2_{FDC}\ge 16S_{AMC}\cdot S_{FNA}$