Kẻ $AH \perp BC;BE \perp AD$
$S_{ABCD}=AH.BC=AD.BE\\ S_{\Delta ABM}=\dfrac{1}{2}AH.BM=\dfrac{1}{2}AH.\dfrac{1}{2}BC=\dfrac{1}{4}S_{ABCD}\\ S_{\Delta ABD}=\dfrac{1}{2}BE.AD=\dfrac{1}{2}S_{ABCD}\\ =>S_{\Delta ABM}=\dfrac{1}{2}S_{\Delta ABD}$
Xét $\Delta MBQ$ và $\Delta ADQ$
$\widehat{M_1}=\widehat{A_1}(MB//AD)$
$\widehat{Q_1}=\widehat{Q_2}(đđ)$
$=>\Delta MBQ$ đồng dạng với $\Delta ADQ$
$=>\dfrac{S_{\Delta MBQ}}{S_{\Delta ADQ}}=(\dfrac{MB}{AH})^2=\dfrac{1}{4}\\ <=>S_{\Delta ADQ}=4S_{\Delta MBQ}\\ S_{\Delta ABM}=\dfrac{1}{2}S_{\Delta ABD}\\ <=>S_{\Delta ABQ}+S_{\Delta MBQ}=\dfrac{1}{2}(S_{\Delta ABQ}+S_{\Delta ADQ})\\ <=>S_{\Delta ABQ}+S_{\Delta MBQ}=\dfrac{1}{2}(S_{\Delta ABQ}+4S_{\Delta MBQ})\\ <=>S_{\Delta ABQ}=2S_{\Delta MBQ}\\ =>S_{\Delta MBQ}=\dfrac{1}{3}(S_{\Delta ABQ}+S_{\Delta MBQ})=\dfrac{1}{3}S_{\Delta ABM}=\dfrac{1}{12}S_{ABCD}\\ =>S_{\Delta ABQ}=\dfrac{1}{6}S_{ABCD};S_{\Delta ADQ}=\dfrac{1}{3}S_{ABCD}\\ S_{MQDC}=S_{ABCD}-(S_{\Delta MBQ}+S_{\Delta ABQ}+S_{\Delta ADQ})=\dfrac{5}{12}S_{ABCD}=\dfrac{5}{12}(m^2)$
