- Membership
- Publications
- Meetings
- Competitions
- Community
- Programs
- Students
- High School Teachers
- Faculty and Departments
- Underrepresented Groups
- MAA Awards
- MAA Grants

- News
- About MAA

Next, “the segment about the base” is the segment bounded by arc AHC and chord AEC. That this segment “is equal to the sum of those about the sides” of triangle ABC follows from a generalization of the Pythagorean theorem, that if similar figures are constructed on the hypotenuse and the two legs of a right triangle, the area of the figure on the hypotenuse is the sum of the areas of the figures on the legs. This theorem can be verified by constructing and measuring these three areas. To construct the segment bounded by arc AHC and chord AEC, select points A, H, and C in that order or in the reverse order. Then select **Arc Through 3 Points** from the **Construct** menu. With arc AHC highlighted, select, from the **Construct** menu, **Construct Arc Interior** and then **Arc Segment**. With the arc segment highlighted, select **Area** from the **Measure** menu. A similar process can be followed to construct and measure the segment bounded by arc AIB and chord AB and by are BJC and chord BC whose areas can then be summed. You can click here to see this construction. Also, see Figure 17.

Figure 17. Arc segments AHC, AIB, and BJC.

We can then use the dynamic capabilities of *Geometer’s Sketchpad* to verify that the segment bounded by arc AHC and chord AEC is the sum of the areas of the segments bounded by arc AIB and chord AB and by are BJC and chord BC. This is done by clicking and dragging one of the two points that define circle E, which is either point E or point A in this example, and noticing that this is always true.

Hippocrates then stated, “it follows that, when the part of the triangle above the segment about the base is added to both” the segment AHCE, which gives the triangle ABC, and the two segments in pink, which gives the lune, “the lune will be equal to the triangle”.

Although this is quite clear from the diagram, we will demonstrate it algebraically as well. Let the length of each leg of triangle ABC equal x. Therefore, the hypotenuse of triangle ABC, which is the diameter of the smaller circle (circle E), is x Ö2. Since the larger circle (circle D) has a diameter of 2x, the ratio of the diameter of the larger circle to the smaller circle is (2x)^{2} /(x Ö2)^{2} = 4x^{2}/2x^{2} = 2/1. Therefore, the area of circle D is twice the area of circle E. It follows that the area of any part of circle D is also twice that of an analogous part of circle E, which proves that the segment on base AEC is double each of the two segments in circle E.

Although Hippocrates fell short of squaring a circle, he was the first to compare the area of a curvilinear figure to that of a rectilinear figure. As a result of squaring the area of a lune, Hippocrates was the first to show that figures with *curved* boundaries could be squared.

Suzanne Harper and Shannon Driskell, "An Investigation of Historical Geometric Constructions - Calculating the Area of a Lune III," *Loci* (August 2010)