Xét hình thang $ABCD\ (AB//CD)$ có:
$AM = MD =\dfrac12AD$
$BN=NC=\dfrac12BC$
$\Rightarrow MN$ là đường trung bình
$\Rightarrow \begin{cases}MN =\dfrac{AB+CD}{2}=\dfrac{3AB}{2}\\MN//AB//CD\end{cases}$
Xét $\triangle ABD$ có:
$AM = MD =\dfrac12AD$
$AP//AB$
$\Rightarrow AP =\dfrac12AB\qquad (1)$
Xét $\triangle ABC$ có:
$BN = NC =\dfrac12BC$
$NQ//AB$
$\Rightarrow NQ =\dfrac12AB\qquad (2)$
Ta lại có:
$\quad MP + PQ + QN = MN$
$\Leftrightarrow PQ = MN - MP - NQ$
$\Leftrightarrow PQ =\dfrac{3AB}{2} -\dfrac12AB -\dfrac12AB$
$\Leftrightarrow PQ =\dfrac12AB\qquad (3)$
Từ $(1)(2)(3)\Rightarrow MP = PQ = QN$