The third type of lune is constructed with a circle circumscribing a square.  We begin with the square ABCD and it's diagonals AC and BD with intersection O1.

Now, we construct a circle whose center is O1 and radius AO1.

Next we find the midpoint, O2, of AB and construct a second circle with center O2 and radius AO2.

Now the lune we are seeking is the area of circle O2 that is not contained within circle O1.

Now we will prove that this yellow lune is in fact quadrable by showing that it is equal in area to triangle ABO1, which we know is quadrable since all triangles are squareable.  Begin by looking at triangle ABC.  We know that angle ABC is a right angle and that AB=BC so we can use the pythagorean theorem to say that (AC)2=(AB)2+(BC)2=2(AB)2.  Now let's look at the ratio of the area of the sector AFB with the area of ABC.  Area(AFB)=(pi)(AB)2 and Area(ABC)=(pi)(2)(AB)2, so the ratio of Area(AFB):Area(ABC)=1:2.  This implies that the area of the quadrant AEBO1 is half the area of the semicircle ABC, therefore the area of the semicircle AFB is equal to the area of quadrant AEBO1.  So we have that Area(AFB)-Area(AEBO1)=Area(AEBO1)-Area(AEBO2).  Therefore Area(LuneAFBE)=Area(TriangleABO1) giving us that the lune is in face quadrable.