M - midpoint of AB, N - of CD. K,L intersection points of line parell to AD through point N and line parell to BC through point N with AB. KLN is right triangle => MK=MN=ML=1/2(AB-CD)

limes123 wrote: M - midpoint of AB, N - of CD. K,L intersection points of line parell to AD through point N and line parell to BC through point N with AB. KLN is right triangle => MK=MN=ML=1/2(AB-CD) Nice and simple solution ! HINT With the above notation (last post), if $ O$ is the common point of $ AD,BC$ , then $ O,M,N$ are collinear and triangle $ OAB$ is right. But $ OM = \frac {AB}{2} , ON = \frac {CD}{2}$ and $ NM = OM - ON$ . Babis XXXXXXXXXXXXXXXXXXXXX

Hi Here's my solution. If $ AB $ and $ CD $ intersect at $ E $ then $ AEB=90 $ Let $ M $ and $ N $ be the midpoint of $ DC $ and $ AB $. Now we have that $ M $ and $ N $ and $ E $ are collinear. So we have $ EM = DC/2 $ and $ EN= AB/2 $ so the problem is solved.

YaMohammad wrote: Hi Here's my solution. If $ AB $ and $ CD $ intersect at $ E $ then $ AEB=90 $ Let $ M $ and $ N $ be the midpoint of $ DC $ and $ AB $. Now we have that $ M $ and $ N $ and $ E $ are collinear. So we have $ EM = DC/2 $ and $ EN= AB/2 $ so the problem is solved. $AB$ is parallel to $CD$

you mean if $AD$ and $BC$ intersect at $E$ ...

this thing could be coordinate bashed

Trapezoid Lemma for 4 collinear points

problem's solution