Answer:
x=45
Explanation:
I can give you two solutions
First solution:
In
we know that
![\angle BFA + \angle FAB + \angle ABF=180\\\\90+x+\angle ABF =180\\\\\angle ABF =90-x](https://img.qammunity.org/2021/formulas/mathematics/high-school/6gwzsuqlobtgb193t05nltrz3vw454u0ze.png)
same case for
in
![\triangle ECB](https://img.qammunity.org/2021/formulas/mathematics/high-school/wg4obsi3uljhvdk09gzd74z3g6giopr7lo.png)
so we have that
![\angle ABC = \angle ABF + \angle FBE +\angle EBC\\\\\angle ABC = 90-x+45+90-x\\\\\angle ABC = 180+45-2x](https://img.qammunity.org/2021/formulas/mathematics/high-school/f6weylpxrzkcno1gc9xa3f621rwkm5e3gm.png)
we leave it like that because it's convenient
Now by properties of parallelograms we know that opposite sides add up to 180
![\angle FAB + \angle ABC = 180\\\\x+180+45-2x=180\\\\x+45-2x=0\\\\45-x=0\\\\x=45](https://img.qammunity.org/2021/formulas/mathematics/high-school/ka07co2uh5ggww5paxtuabs14d729rc201.png)
so one angle is 45 and the other one is 135
Solution 2
if we look at DEBF we can say that it's a cyclic quadrilateral
because opposite sides add up to 180
in our case
![\angle DFB = 180 -\angle BFA = 180-90 = 90\\\\\angle DEB = 90\\\\\Rightarrow \angle DFB + \angle DEB = 180\\\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/8rxrbhujz8curynp8yfhoooy9zl5dd0xh6.png)
so for the other to angles is also true
![\angle FDE + \angle EBF = 180\\\\\angle FDE +45 = 180\\\\\angle FDE = 135](https://img.qammunity.org/2021/formulas/mathematics/high-school/srxtr1duehqarbx89s3pxovblrroaidyiz.png)
so that's the measure of one angle of the parallelogram
the other angle x is
![\angle EDF + \angle FAB = 180\\\\135+x=180\\\\x=45](https://img.qammunity.org/2021/formulas/mathematics/high-school/vc8d6ot9l2scivi5xo3cho1gq6g2zongk8.png)
There you go choose yourself