|In right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle.|
|Let ABC be a right-angled triangle having the angle BAC right.
I say that the figure on BC equals the sum of the similar and similarly described figures on BA and AC.
|Draw the perpendicular AD.||I.12|
|Then, since in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, therefore the triangles DBA and DAC adjoining the perpendicular are similar both to the whole ABC and to one another.||VI.8|
|And, since ABC is similar to DBA, therefore BC is to BA as BA is to BD.||VI.Def.1|
|And, since three straight lines are proportional, the first is to the third as the figure on the first is to the similar and similarly described figure on the second.||VI.19,Cor|
|Therefore BC is to BD as the figure on BC is to the similar and similarly described figure on BA.|
|For the same reason also, BC is to CD as the figure on BC is to that on CA, so that, in addition, BC is to the sum of BD and DC as the figure on BC is to the sum of the similar and similarly described figures on BA and AC.||V.24|
|But BC equals the sum of BD and DC, therefore the figure on BC equals the sum of the similar and similarly described figures on BA and AC.|
|Therefore, in right-angled triangles the figure on the side opposite the right angle equals the sum of the similar and similarly described figures on the sides containing the right angle.|
The broad problem Hippocrates was investigating was that of quadrature of a circle, also called squaring a circle, which is to find a square (or any other polygon) with the same area as a given circle. Hippocrates did not solve that problem, but he did solve a related one involving lunes. A lune (also called a crescent) is a region of nonoverlap of two intersecting circles. Hippocrates did not succeed in squaring an arbitrary lune, but he did succeed in a couple special cases. Here is a summary of the simplist case.
|Draw a square ABCD with diameters AC and BD meeting at E. Circumscribe a semicircle AGBHC about the right isosceles triangle ABC. Draw the arc
AFC from A to C of a circle with center D and radius DA.
Hippocrates finds the area of the lune formed between the semicircle AGBHC and the arc AF as follows.
Note that there are three segments of circles in the diagram; two of them are small, namely, AGB with base AB, and BHC with base BC, and one is large, namely, AFC with base AC. The first two are congruent, and the third is similar to them since all three are segments in quarters of circles.
Therefore, the lune, which is the semicircle minus the large segment, equals the semicircle minus the sum of the small segments. But the semicircle minus the sum of the small segments is just the right triangle ABC. Thus, a rectilinear figure (the triangle) has been found equal to the lune, as required.
Note that at Hippocrates' time, Eudoxus' theory of proportion had not been developed, so the understanding of the theory of similar figures (Book V) was not as complete as it was after Eudoxus. Also, Eudoxus' principle of exhaustion (see Book XI and proposition X.1) for finding areas of curved figures was still to come, so the concept of area of curved figures was on shaky ground, too. Such a situation is common in mathematicsmathematics advances into new territory long before the foundations of mathematics are developed to logically justify those advances.
Next proposition: VI.32
© 1996, 2002