Let $E$ be a point outside square $ABCD$ such that $AE=AC$, $F$ be a point on $AE$ such that $DE=DF$, and $G$ be the intersection point of $AC,BF$ and satisfy $BG=EC$.Prove that $BG||EC$.
This conclusion is easy to verify by constructing figure. And, we can see $\angle EAC=30^o$. But how to prove it by pure geometric method only? THX in advance.
This is probably not a 'pure' geometrical proof but may be of some interest.
As you have pointed out the configuration, C say, with $\angle BAG=45^o$, $\angle ABG=30^o$ is easy to find so all that is necessary is to prove there are no other solutions.
Let $AD$ have unit length and, initially, ignore the condition $BG=EC$. Then the locus of $E$ is a circle centre $A$.
As $E$ moves anti-clockwise from $C$ until it reaches the line $AD$, the mid-point of $E$ and $F$ moves on a circle with $AD$ as diameter and so $F$ moves on a limacon (the relevant part of which is part of a simple convex loop with $AD$ as a line of symmetry) from $A$ until it reaches the line $AD$ again (at distance $(2-\sqrt 2)$ from $A$).
During this 'motion', $|CE|$ increases from $0$ while $|BG|$ decreases from $1$. These lengths become equal at position C. In this configuration $BG$ intersects the limacon at $15^o$ to its tangent and intersects $AD$ inside the limacon. Therefore there is only one position where $BG=CE$.