Prove the given geometry problems. Problem 1: In ABC, AE=EB and EF||BC. In ACD, FG||CD. Prove AG=GD. Problem 2: In DEF, DS=SE, EU=UF, and ST||EF. Prove STUE is a parallelogram.

We are given that $AE = EB$, which means $E$ is the midpoint of side $AB$.
We are also given that $EF \parallel BC$.
By the converse of the Midpoint Theorem (also known as Thales's Theorem or Basic Proportionality Theorem), if a line parallel to one side of a triangle passes through the midpoint of another side, then it bisects the third side.
Therefore, $F$ must be the midpoint of side $AC$. This implies $AF = FC$.

From Step 1, we established that $F$ is the midpoint of side $AC$.
We are given that $FG \parallel CD$.
Applying the converse of the Midpoint Theorem to $\triangle ACD$:
Since $FG \parallel CD$ and $F$ is the midpoint of $AC$, then $G$ must be the midpoint of side $AD$.
Therefore, $AG = GD$.

This completes the proof.

:
Given:
In $\triangle DEF$, $DS = SE$, $EU = UF$, and $ST \parallel EF$.
Prove: $STUE$ is a parallelogram.

We are given that $DS = SE$, which means $S$ is the midpoint of side $DE$.
We are also given that $ST \parallel EF$.
By the converse of the Midpoint Theorem, if a line parallel to one side of a triangle passes through the midpoint of another side, then it bisects the third side.
Therefore, $T$ must be the midpoint of side $DF$.

Since $S$ is the midpoint of $DE$ and $T$ is the midpoint of $DF$, by the Midpoint Theorem, the segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
So, $ST = \frac{1}{2} EF$.

We are given that $EU = UF$, which means $U$ is the midpoint of side $EF$.
Therefore, $EU = \frac{1}{2} EF$.

From Step 2, $ST = \frac{1}{2} EF$.
From Step 3, $EU = \frac{1}{2} EF$.
Thus, $ST = EU$.

We are given that $ST \parallel EF$.
Since $U$ lies on $EF$, it follows that $ST \parallel EU$.

We have shown that in quadrilateral $STUE$, one pair of opposite sides ($ST$ and $EU$) are both parallel ($ST \parallel EU$) and equal in length ($ST = EU$).
A quadrilateral with one pair of opposite sides that are both parallel and equal in length is a parallelogram.
Therefore, $STUE$ is a parallelogram.

This completes the proof.