As shown in the figure below: (please scroll down a bit to see the figure) AB’ is one diameter of the circle, and a blue curve from A to B is supposed to bisect the area of the circle. We see from the figure below: 
Length of the blue curve AB is greater than: AE + EB = AE + EB’
which is certainly longer than AB’; AB’ is the diameter.
The blue curve is obvious the focus. (Note when blue curve AB is mentioned, we mean the “curved one”: it curves around, not going straight from A to B.) Besides, point E is where the blue curve intersects with CD; and CD is one diameter.
So we have proved the claim that any curve bisecting the circular area got to be longer than the diameter.
Stay with us for one more minute. Let the construction-proof process be revealed to you, as follows.
Referring back to the figure. Let us start from the circle and the blue curve only (imagine all other lines and points disappear; now we draw them step by step). Connect the two endpoints A, B of the blue curve by a line segment AB. Then draw the diameter CD // AB (i.e. line CD is parallel to line AB).
Take point O (the midpoint of CD), and then passing A and O, let another diameter AB’ be drawn.
For completing the proof , two arguments are required:
(1) The blue curve has at least one intersection point with diameter CD (thinking it: if the blue curve resides at only one side of CD, then that curve cannot divide the circular area evenly into two parts with equal area; therefore, any curves that bisects the circular area must intersects diameter CD). Now suppose the intersection point is E.
(2) B and B’ are symmetric to the diameter CD therefore EB = EB’ (trying justifying it using the property of circles and parallel lines).
The rest of the proof is straightforward (and intuitive).