Let $M,N$ be midpoints of $AD,BC$. Then $M,E,N$ are collinear. Suppose $\angle AED$ is acute. Then $2MN>AD+BC$. But since $AB+DC\geq 2MN$, we have $AB+DC>AD+BC$, a contradiction, since $AB+DC=AD+BC$. So $\angle AED$ is not acute.

oneplusone wrote: $2ME>AD+BC$. But since $AB+DC\geq 2MN$ ... Should that be $2MN>AD+BC$? And please could you explain why this and the following inequality$AB+DC\geq 2MN$ hold? thanks Merlin

In any triangle, we have $b+c \geq 2m_a$, where $m_a$ is the $A$-median. To see this, reflect $A$ over the midpoint of $BC$ to get $A'$ and use triangle inequality in $\triangle AA'B$. Here after translating $MN$ to $M'B$ and $CD$ to $BD'$ we can directly use this result in $\triangle ABD'$ with median $AM'$.

Merlinaeus wrote: oneplusone wrote: $2ME>AD+BC$. But since $AB+DC\geq 2MN$ ... Should that be $2MN>AD+BC$? And please could you explain why this and the following inequality $AB+DC\geq 2MN$ hold? thanks Merlin Yeah, edited it should be $MN$. This is because $2ME>AD$, since $E$ lies outside the circle with diameter $AD$. Similarly $2NE>BC$. For $AB+DC\geq 2MN$ is just as Kenny_cz said.