The Pentagon and the Star Pentagon

The sides of the Pentagon and the Star Pentagon, or the chord subtended by 2/5 of the Circle, have a Geometrical description through their angles, and are knowable, separately in the 8th degree; combined, in the sixth and in the fourth degree of knowledge.

Description independently of the circle proceeds thus: if the proposed (futurum) side is given in length, we shall divide it in proportional section by Euclid II. 11 or VI. 30, and to it we shall join the greater part formed by the section: and having drawn two sides (crura) each equal to this composite line; and making the proposed line the Base, we shall construct the interior triangle of the Pentagon.

The composite side (crus) comprises the whole proposed line and the greater part is found hy dividing it in the divine section.

The composite line will also be divided in this way [ie. in proportional section], and its greater part will be the proposed side, so that the base angle of this Triangle will be twice the angle at the vertex, as above for the Decagon and on the two said equal sides (crura) of the triangle, serving as bases, we shall add two triangles to the outside [of the figure], the triangles having their [remaining] sides (crura) equal to the proposed side.

The easiest inscription in the circle is via the side of the Decagon. For, since half of ten is five, if we join up the ends F, H o f two sides FG, GH of the Decagon that meet at G, the line FH will be the side of the Pentagon, and similarly for and if we join up the ends F, K, the line FK will be the side of the star.

So let the Pentagon be BDFHK and its star BFKDHB.

Euclid, then, shows in 13. 10 that the square of FH, the side of the Pentagon, is equal to the sum of the square of FA, the side of the Hexagon, and of the square of FG, the side of the Decagon, that is [the sum of the squares of] the semidiameter, AG, and GO, the Greater part resulting from the [Proportional] section [of the semidiameter].

This proof in Euclid is somewhat dijficult to understand.

So I shall try to give an easier one here.

From the ends of the side of the Pentagon B, D let there be drawn through the center A the straight lines BG and DI: and as DB subtends two tenths [of the circle], similarly let the neighboring line DL subtend three [tenths] and DK four [tenths], these lines cutting BG in the points S and R [respectively[.^’^^ So the angle EDI, that is SDA, is two fifths of a Right angle, because [the arc] L I is one fifth part of the circle, just as FH also, and indeed arcs equal to it, subtend equal angles at the circumference, by Euclid III.21 or 27.^"^'^

Angle DAB, that is angle DAS, is equal to four fifths of a right angle, because DB is a fifth part of the circle, whose complete circumference marks out four right angles at A.

So the sum of angles SAD and ADS is six fifths of a right angle. But all three [angles of the triangle SAD] add up to ten fifths.

Therefore the remaining angle, DSA, is four fifths. So [angle] DSA is equal to the [angle] DAS, and the side DS is thus equal to the side DA, which is a semidiameter. Therefore, by the above, the greater part obtained by proportional section of the semidiameter DA is equal to SA, so SA is equal to the side of the Decagon, by what has [already] been said.

DA is the semidiameter, that is the side of the Hexagon. So I say that the side of the Pentagon, DB, squared, is equal to the sum of the squares of SA and AD.

For, joining K to S and to A, since DA, AK are equal, and DS, SK are equal to them, the parts SR, RA will also be equal, and angle DRB is a right angle.

Therefore DB squared is equal to the sum of DR squared and RB squared.

But DR squared is less than DA squared, by the amount RA squared, and BR squared is less than BA squared by an amount which is the sum of the rectangle contained by BR, RA, taken twice, and the square of the line RA.

So the sum of the squares of DR and RB is less than the sum of the squares of DA and AB by twice the rectangle contained by SA, AB, that is by the rectangle RA, AB taken once.

But the two rectangles contained by SA, AB and SB, BA together make up the whole square of BA.

Therefore, on subtracting the rectangle contained by SA, AB, there remains the square of the line DA, plus the rectangle contained by SB, BA, and together they are equal to the square of DB.^^^ Now, since the semidiameter, BA, is divided in proportional section at S, and the greater part is AS: so the rectangle SB, BA is equal to the square of Therefore the side of the Pentagon, squared, is equal to the [sum of the] two squares of DA and AS; that is the squares of the sides of the Hexagon and the Decagon.

Regarding the side of the star Pentagon BF: this is composed from BD, or BQ the side of the Pentagon, and from QF, the greater part derived from it by proportional section: by [Euclid] XIII.8^^^: which result can also be proved from the triangle [used in the construction] of the five-cornered figure, [triangle] FBII, as above.

So since the square of the side of the Pentagon is equal to the square of the semidiameter, which is expressible in length, plus the square of the Greater part derived from it by proportional section, as in the diagram of the semicircle above, PG squared is equal to the sum of PA squared and AG squared, and the ratio ofPA to AG will be equal to the ratio ofPG, the side of the Pentagon, to the side of its star: indeed the ratio of PA to AG will be equal to that of PG to Therefore GX is the side of the star, and its square is the sum of the square of GA, the semidiameter of the circle surrounding the ten-angled figure, and of the square of the line AX, composed of PA and AG. Thus, by what is proved there,^^^ GX is a Mizon, GP an Elasson.

Individually they belong to the eighth degree of knowledge, and to its second level (ordine). Because, taken together the lines PG, GX make the sum of their squares Expressible, namely equal to the square PX, which is five times the square of the Expressible line GA.

the same two lines PG, GX give a Medial rectangle. On this basis PG, GX, taken together, belong to the 6th degree of knowledge which was discussed in Part 18.

The side of the Pentagon and the side of the star are related as the larger part and the whole in the divine section. They accordingly also belong to the 4th degree of knowledge, when combined with one another: see Part 29.

It follows from these properties that just as the side of the Pentagon is an Elasson, and that of the Star a Mizon, so too the line composed of them will again be a Mizon, and the side of the Pentagon will be the smaller element of this compound line, considered as a Mizon; while the side of the star will be its greater Element; and similarly also, the difference between the two sides will be an Elasson, that is Z)Q or QF, by Part 29.