Sum of angles??

Sum of angles:

$4\times 180^\circ - 360^\circ =720^\circ - 360^\circ =360^\circ$ ,

Keep in mind that sum of angles in a quadrilateral is $A+B+C+D=360^\circ$ and sum of angles in a triangle is $180^\circ$

EDIT : Acknowledging @Steven Adler 's comment that the central figure in this problem is not a parallelogram but rather, a quadrilateral. This is to avoid confusion from anyone. Forgive me for late edit in the explanation.

We know that the sum of all interior angles of any quadrilateral amounts to $360^{\circ}$ . So, what I did in my solution was to cut down the quadrilateral into two triangles for simplicity's sake. We then know that all triangles have interior angles summing up to $180^{\circ}$ . We can now use this information to find the sum of all angles colored orange from the figure above.

Let's also take the fact that vertically opposite angles are equal. This means that each $\color{#EC7300}{orange}$ indicated angles of the outer triangles equals the $\color{#3D99F6}{blue}$ angles (or corners) of the quadrilateral divided into two triangles as shown below.

So, the sum of given angles will be:

$(2\times180^{\circ})+(2\times180^{\circ})-360^{\circ}=\boxed{360^{\circ}}$

The sum is 360 degrees for ANY polygon with a triangle at each vertex.

For any N-gon:

There are N surrounding triangles.

Sum of all triangle angles is 180N

Subtract sum of all interior angles of N-gon, 180(N-2). These are the vertical angles subtracted from each surrounding triangle.

180N - 180(N-2) = 180N - 180N + 180x2 = 360

QED