Giải thích các bước giải:

a.Vì $ABCD$ là hình bình hành $\to AD//BC, AB//CD$

$\to \dfrac{MD}{MK}=\dfrac{MA}{MC}=\dfrac{MN}{MD}$

$\to MD^2=MK\cdot MN$

b.Ta có:

$DM\cdot (\dfrac1{DN}+\dfrac1{DK})$

$=\dfrac{DM}{DN}+\dfrac{DM}{DK}$

$=\dfrac{CM}{CA}+\dfrac{AM}{AC}$

$=\dfrac{CM+AM}{AC}$

$=\dfrac{AC}{AC}$

$=1$

$\to \dfrac1{DN}+\dfrac1{DK}=\dfrac1{DM}$