prove that the quadrilateral $ABCD$ is a square

Question

Given $ABCD$ a quadrilateral such that $AB\parallel CD$ and $\angle ACD=45^0, \angle A=90^0, \angle D=90^0 $

Need to prove that $ABCD$ is a square.

I tried to use circles but it didn't help.

Any ideas?

thanks.

EDIT: i added a crucial details about angles $A$ and $D$.

Answer 1

Its not a square. You are given that $\angle A$ and $\angle D$ are $90^\circ$, but you are not guaranteed in $BC \parallel AD$. In fact, $\angle D = 90^\circ$ is implied by $AB \parallel CD$