In the adjoining figure, $ABCD$ is a parallelogram In the adjoining figure, $ABCD$ is a parallelogram. If $DF=CE$ and $AG||HF$ then prove that $||gm FAGH =||gm ABCD$. !enter image description here My Attempt. $1. FD=EC$ $2. FD+DE=EC+DE$ $3. FE=DC$. What should I do further to complete this?

From $CE=DF$, we have $$[\triangle{BEC}]=[\triangle{AFD}]\tag1$$

Also we get $$\begin{align}AB&=CD\\\&=CE+ED\\\&=DF+ED\\\&=FE\end{align}$$ so $ABEF$ and $FAGH$ are parallelograms.

Since $HF=GA,FE=AB$ and $\angle{HFE}=\angle{GAB}$, we get $$\triangle{HFE}\equiv \triangle{GAB}$$ from which we get $$[\triangle{HFE}]=[\triangle{GAB}],$$ i.e. $$[\square{HGDF}]+[\triangle{GED}]=[\square{DEBA}]+[\triangle{GED}]$$ So, $$[\square{HGDF}]=[\square{DEBA}]\tag2$$

The claim follows from $(1)$ and $(2)$.