Elementary geometrical proof involving a trapezoid

Question

In a trapezoid ABCD is M = mi $[DB]$ and N= mi $[AC]$

Proof that $|MN| = \frac {1}{2} (|DC| - |AB|)$

I've already proven this analytically with points A(2,4); B(6,4); C(10,0); D(0,0) and in the general sense just by using slopes. But I don't know how to proof this geometrically.

Can someone give me a hint please?

Answer 1

Let $P$ and $Q$ be the intersections of $MN$ with $AD$ and $BC$.
By Thales' theorem $MN\parallel AB\parallel CD$ and $$ PM=NQ=\frac{1}{2}CD,\qquad PQ=\frac{AB+CD}{2} $$ hence it follows that: $$ MN = PQ-PM-NQ = \frac{AB+CD}{2}-CD = \color{red}{\frac{AB-CD}{2}} $$ as wanted.