Hero of Alexandria, Metrica i Preamble, 1-9

translated by Henry Mendell (Cal. State U., L.A.)

Return to Vignettes of Ancient Mathematics

The text used is Hero Alexandrinus,

Operaiii, ed. Hermann Schöne (Leipzig: Teubner, 1903), abbrev., Sch.

Prop. 2: the area of a right triangle

Prop. 3: the area of an isosceles triangle

Prop. 4: the area of a non-isosceles triangle (general geometrical remarks)

Prop. 5: the area of an acute-angled triangle

Prop. 6: the area of an obtuse-angled triangle

Props. 7: a theorem on ‘sides of squares’ (used in Prop. 8), preceded by an introduction to Prop. 8

Prop. 9: the area of a triangle where the altitude is not rational.

This text uses unicode. The following translations have been abopted:

ἐμβαδόν = area

χωρίον = plot

μέθοδος = procedure

Heron mostly uses Egyptian fractions (standard in Greek literature from about 300 BCE). These are so-called unit fractions. I shall use:

∠ = ½

*n*´ = *n*’th marker (e.g, 4´ = ¼)

A number such as 1 5/6 might be written: 1
∠
3´ (1 + ½ + 1/3)

Operations

multiplication: x times y (this will translate the Greek, τὰ x ἐπὶ τὰ y)

subtraction: y is taken away from x

squaring: x times itself (τὰ x ἐφ᾽ ἑαυτά)

square root: side of square x. The term is geometrical in origin, but is used in a numerical sense and may be used geometrically

Geometrical configurations

square: the square from AB (abbrev: that from AB)

rectangle: the rectangle enclosed by ΑΒ, ΒΓ (abbrev: that by ΑΒ, ΒΓ; abbrev: that by ΑΒΓ)

angle: the angle enclosed by ΑΒΓ (abbrev: that by ΑΒΓ; but usally, angle ΑΒΓ, to disambiguate from ‘rectangle’, where the Greek is clear since for ‘rectangle’, the definite article (trans. “that’ or ‘what's’) is neuter, while for ‘angle’, it is feminine.

The first geometry, as the ancient story teaches us, engaged in measurements and divisions on land, whence it was also called ‘geometry (land or earth measure)’.1 Since the subject matter was useful to people, its sort (γένος) made much progress, so that the management of measurement and divisions also proceeded to solid bodies. And since the first theorems conceived (ἐπινοηθέντα) were not adequate, they required further more extraordinary investigation so that even up to now some of them raise difficulties, although Archimedes and Eudoxus nobly contributed to the subject. For there was no means of producing a demonstration before the conception (ἐπίνοια2) of Eudoxus through which the cylinder is three-times the cone having the same base and height equal to it (Eucl., *El*. xii 10, cf. Arch., *De sphaer. et. cyl*., intro., and *Meth*., intro.), and that circles are to each other as the squares from the diameters are to one another. And it was incredible to conceive before the apprehension of Archimedes, why the surface of the sphere is four-times the largest of the circles in it (Arch., *De sphaer. et cyl*. i 33), and that its volume is two third-parts of the cylinder circumscribing it (Arch., *De sphaer. et cyl*. i 34), and any siblings of these. And so, the mentioned subject is required, we consider that it would be well to collect also all we have additionally observed. Let us begin from the measurements of planes, including along with planes also the other surfaces, convex and concave, since every surface is conceived from two extensions. And the comparisons of the mentioned surfaces come-about in relation to some rectilinear and rectangular plot: rectilinear, since the straight-line is also invariable in comparison with other lines, since every straight-line fits on every straight-line,while the others, convex and concave lines, do not all of them fit on all of them. [lacuna? Cf. Sch.]Wherefore with regard to something fixed, I mean the straight line, they would furthermore make the comparison with the right angle.3 For again every right angle fits on every right angle, while the others do not all of them fit on all of them. A cubit area is called when a square plot has each side of one cubit. Similarly, a foot area is called, when a square plot has each side of one foot. Thus the mentioned surfaces have comparisons with the mentioned plots or the parts of them. Again, solid bodies then get their relations to a solid, rectilinear and rectangular plot, everywhere equilateral. This is a cube having each side of either one cubit or one foot. Or again, with the parts of these. And so, it has been stated the reason why the comparison comes-to-be with the mentioned plots, but let us begin in succession with measurements on surfaces. And so, in order that we don't name in each measurement feet or cubits or the parts of this, we will display the numbers in units, since it is possible to suppose them with respect to whatever measure is intended.4

This is a variation on a plausible story. Cf. Herodotus, *Hist*. 2.109.

One might think that Hero's frequent use of ‘conceive (ἐπίνοια)’, a technical term in Stoicism (attending the mind to an object of thought), reveals him a Stoic. It is better to assume that Hero just chooses a ordinary terminology of his time.

Hero's reasoning is a little odd (every 30˚ angle fits on every other) unless one thinks of the species of rectilinear angles as acute, right, obtuse.

Heron's use of units (μονάδες) as an abstract measure departs from the standard practice in the mensuration literaure (Babylonian, Egyptian, and Greek).

Prop. 1

(diagram 1)

1. Let there be an oblong plot ΑΒΓΔ, havingΑΒ as 5 units and ΑΓ as 3 units. To find its area. Since every right-angled parallelogram is said to be enclosed by two straight-lines enclosing a right angle, and what’s enclosed by ΒΑ, ΑΓ is of this sort, the area of the oblong will be 15 units. (diagram 2) For if each side is divided, ΑΒ into 5 units, and ΑΓ similarly into 3 units and through the sections parallels are drawn to the sides of the parallelogram, the plot will be divided into 15 plots, of which each will be 1 unit. And if the plot is a square, the same argument (λόγος) will apply.

Prop 2

(diagram 1)

2. Let there be a right-angled triangle, ΑΒΓ having as right the angle at Β. And let ΑΒ be 3 units, and ΒΓ 4 units. To find the area of the triangle and the hypotenuse. (diagram 2) Let right-angled parallelogram ΑΒΓΔ be filled out, where the area, as was shown above, is 12 units. But triangle ΑΒΓ is half of parallelogram ΑΒΓΔ. And so, the area of triangle ABG will be 6 units; (diagram 3) and since the angle at B is right, the squares from ΑΒ, ΒΓ are equal to the square from ΑΓ. And the squares from ΑΒ, ΒΓ are 25 units. And that from ΑΓ is, therefore, 25 units. ΑΓ itself is, therefore, 5 units.

The procedure is this: making 3 times 4 to take half of these. They become 6. The area of the triangle is so much. …having made 3 times themselves and similarly 4 times themselves, to add them. And they become 25. And having taken a side of these to have the hypotenuse of the triangle.

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Prop 3

(general diagram = diagram 4)

(diagram 1) 3. Let there be an isosceles triangle, ΑΒΓ, having ΑΒ equal to ΑΓ, and each of the equal units 10, while ΒΓ {to ΑΓ and each of the 10 equal units}^{1}
of units 12. To find its area. (diagram 2) Let an altitude, ΑΔ, be drawn to ΒΓ. (diagram 3) And through Α, let a parallel to ΒΓ be drawn, i.e., ΕΖ, while through Β, Γ let parallels to ΑΔ, be drawn, i.e., ΒΕ, ΓΖ. Therefore, parallelogram ΒΓΕΖ is double triangle ΑΒΓ. For it has the same base as it and is in the same parallels. And since it is isosceles and ΑΔ was drawn perpendicular, ΒΔ is equal to ΔΓ. And ΒΓ is of 12 units. Therefore, ΒΔ is of 6 units. But ΑΒ is of 10 units. (diagram 4) Therefore, ΑΔ is of 8 units, since the square from ΑΒ is equal to those from ΒΔ, ΔΑ, so that also ΒΕ will be of 8 units. But ΒΓ is of 12 units. Therefore, the area of parallelogram ΒΓΕΖ is of 96 units. Thus the area of triangle ΑΒΓ is of 48 units. And this is the method. Take half of 12. It becomes 5. And 10 to themselves. They become 100. Take away 6 to themselves, which are 36. They become remainders 64. The side of this becomes 8. Altitude ΑΔ will be so much. And 12 against 8. They become 96. The half of these. They become 48. The area of the triangle is of so many units.

1This is clearly a dittography (i.e., where the copyist writes the same phrase twice).

Of non-isosceles triangles it is required to observe the angles so that we see the altitudes from the drawn angles to the sides, whether they fall within the angles or outside. And so, let there be a given triangle, ΑΒΓ, having each side of given portions (μοιρῶν). And it is required to observe, as it happens, the angle at Α, whether right or obtuse or acute. And so, if the square from ΒΓ is equal to the squares from ΒΑ, ΑΓ, it is clear that the angle at Α is right, while if it is smaller, acute, and if larger, it is clear that the angle at Α is obtuse. Let it be supposed, in fact, that the square from ΒΓ is smaller than the squares from ΒΑ, ΑΓ. Therefore, the angle at Α is acute. For if it is not acute, it is either right or obtuse. And so, it is not right. For the square from ΒΓ would have to be equal to the squares from ΓΑ, ΑΒ. But it is not. Therefore, the angle at Α is not right. However, it is not obtuse either. For the square from ΒΓ would have to be larger than the squares from ΓΑ, ΑΒ. But it is not. Therefore, it is not obtuse. But it was shown that it is not right either. Therefore, it is acute. Similarly, in fact, we will also conclude that if the square from ΒΓ is larger than the squares from ΒΑ, ΑΓ the angle at Α is obtuse.

Prop. 5

(diagram 1) Let there be an acute-angled triangle, ΑΒΓ having side ΑΒ of 13 units, ΒΓ of 14 units, and ΑΓ of 15 units. To find its area. It is obvious that the angle at Β is acute. For the square from ΑΓ is smaller than the squares from ΑΒ, ΒΔ. (diagram 2) Let a perpendicular, ΑΔ be drawn to ΒΓ. Therefore, that from ΑΓ is smaller than the squares from ΑΒ, ΒΓ by twice the rectangle enclosed by ΓΒ, ΒΔ, as has been shown (El. ii 13). And those from ΑΒ, ΒΓ are of 365 units, that from ΑΓ of 225 units. Therefore, twice the rectangle enclosed by ΓΒ, ΒΔ is of 140 units. Therefore, once that by ΓΒ, ΒΔ will be of 70 units. And ΒΓ is of 14 units. Therefore, ΒΔ will be of 5 units. And since that from ΑΒ is equal to those from ΑΔ, ΔΒ, and that from ΑΒ is of 169 units, while that from ΒΔ is of 25 units, therefore, a remainder, that from ΑΔ, will be of 144 units. Therefore, this ΑΔ will be of12 units. But ΒΓ is also of 14 units. Therefore, that by ΒΓ, ΑΔ will be of 168 units. And it will be double triangle ΑΒΓ. Therefore, triangle ΑΒΓ will be of 84 units.

The procedure will be like this: 13 times themselves. They become 169. And 14 times themselves. they become 196. And 15 times themselves. They become 225. From these take away 224. They become remainders 140. Half of these. They become 70. Apply along 14. They become 5. And 13 times themselves. They become 169. That from which 5 times themselves were taken away. Remainders 144. The side of these become 12. The perpendicular will be of so much. Multiply these times 14. They become 168. Half of these is 84. The area will be of so much.

(diagram 1) 6. Let there be an obtuse-angled triangle, ΑΒΓ having side ΑΒ of 13 units, ΒΓ of 11 units, and ΑΓ of 20 units. To find its perpendicular and area. (diagram 2) Let ΒΓ be extended and let an altitude, ΑΔ, be drawn to it. Therefore, that from ΑΓ is larger than those from ΑΒ, ΒΓ by twice [the rectangle enclosed] by ΒΓ, ΒΔ (*El*. ii 12). And that from ΑΓ is of 400 units, while that from ΒΓ is of 121 units, and that from ΑΒ is of 169. Therefore, what’s twice by ΓΒ, ΒΔ is of 110 units. Therefore, what’s once by ΓΒ, ΒΔ is of units 55 and ΒΓ is of units 11. Therefore, ΒΔ will be of 5 units. But also ΑΒ is of 13 units. Therefore, ΑΔ will be of 12 units. But also ΒΓ is of 11 units. Therefore, that by ΑΔ, ΒΓ will be of 132 units. And it is double triangle ΑΒΓ. Therefore, triangle ΑΒΓ will of 66 units.

The procedure will be this. 13 times themselves become 169. And 11 times themselves become 121. And 20 times themselves become 400. Add 169 and 121. They become 290. Take these from 400. Remainders 110. Have of these become 55. Apply along 110. They become 5. And 13 times themselves become 169. Take 5 times themselves. Remainders 144. The side of these become 12. These times 11 become 132. Half of these is 66. The area of the triangle will be of so much.

topAnd so, up to this point we have make the geometrical demonstrations by calculating, but in the following we will make the measurements following an analysis through the synthesis of numbers.

7. If there are two numbers ΑΒ, ΒΓ, the number enclosed by ΑΒΓ will be the side of the square from ΑΒ times the square from ΒΓ. For since as ΑΒ is to ΒΓ, so is the square from ΑΒ to the number contained by ΑΒΓ and the number contained by ΑΒΓ to the square from ΒΓ, therefore too, as the square from ΑΒ will be to the number contained by ΑΒΓ, so will be the number contained by ΑΒΓ to the square from ΒΓ. And so since three numbers have proportion, the number contained by the extremes is equal to the square from the middle. Therefore, the square from ΑΒ times the square from ΒΓ is equal to the number contained by ΑΒΓ times itself. Therefore, the number contained by ΑΒΓ is the side of the square from ΑΒ times the square from ΒΓ.

Let the numbers be *a*, *b*

*a* · *b* =
$\sqrt{\mathrm{a2\xb7b2}}$

proof: *a* : *b* = *a*^{2} : *a* · *b* and *a* : *b* = *a* · *b* : *b*^{2}

therefore, *a*^{2} : * a* · *b* = *a* · *b* : *b*^{2}

therefore, *a*^{2} · *b ^{2}* = (

There is a general procedure for finding the area given three sides of any triangle whatever and without the altitude. For example, let the sides of the triangle be of 7, 8, 9 units. Compose 7 and 8 and 9. They become 24. Take half of these. They become 12. Take away the 7 units. Remainders 5. Again take away the 8 from the 12. Remainders 4. And further the 9. Remainders 3. Make 12 times 5. They become 60. These times 4. They become 240. These times 3. They become 720. Take the side of these and it will be the area of the triangle.

And so, since 720 do not have its side as rational, we will take the side with a least difference in this way. Since a square near to 720 is 729 and has a side 27, partition 720 into 27. They become 26 and two thirds. Add on 27. They become 53 two thirds. Half of these become 26 ∠ 3´. For 26 ∠ 3´ times themselves become 720 36´. Thus, the difference is 36´ portion of a unit. If we want the difference to come-to-be in a smaller portion than 36´, instead of 729 we arrange the presently found 720 and 36´, and by doing the same things, we will find the difference becoming much smaller than 36´.

Notes: The procedure for the area is: Let the sides of the triangle be *a*, *b*, *c*, and the perimeter *p* = *a* + *b* + *c*. Then, the take *p*/2 · (*p*/2 - *a*) · (*p*/2 - *b*) · (*p*/2 - *c*). Observe that in Greek terms, this is the square of the area. See how this works out in the demonstration. The area of the triangle is then the side of this:
$\sqrt{\mathrm{p/2\; \xb7\; (p/2\; -a)\; \xb7\; (p/2\; -b)\; \xb7\; (p/2\; -c)}}$
.

Heron also provides a method for approximating square roots. Suppose we want
$\sqrt{n}$
. Let
$\sqrt{\mathrm{p2}be\; near}$*n. *
Take 1/2(*p*+*n*/*p*) as a first approximation, and proceed if desired. Note that (1/2(*p*+*n*/*p*))^{2} = 1/4 (*p*^{2} + 2 *n + *(*p/n*)^{2}), so that 1/4 (*p ^{2}*+ 2

Here is the geometrical demonstration of this: the sides of a triangle being given, to find the area. And so, it is possible by drawing a single altitude of it and providing its magnitude to find the area of the triangle, but let it be required to provide the area without the altitude.

(diagram 1) Let the given triangle be ΑΒΓ, and let each of ΑΒ, ΒΓ, ΓΑ be given. To find the area. (diagram 2) Let a circle, ΔΕΖ, whose center is Η, be inscribed in the triangle (*El*. iv 4), and let ΑΗ, ΒΗ, ΓΗ, ΔΗ, ΕΗ, ΖΗ be joined. Therefore, that by ΒΓ, ΕΗ is double triangle ΒΗΓ, while that by ΓΑ, ΖΗ is double triangle ΑΓΗ, and that by ΓΑ, ΖΗ is double triangle ΑΓΗ. Therefore, that by the perimeter of triangle ΑΒΓ and ΕΗ, that is that from the center of circle ΔΕΖ, is double triangle ΑΒΓ. (diagram 3) Let ΓΒ be extended, and let ΒΘ lie equal to ΑΔ. Therefore, ΓΒΘ is half the perimeter of triangle ΑΒΓ, due to the fact that ΑΔ is equal to ΑΖ and ΔΒ to Β and ΖΓ to ΓΕ. Therefore, that by ΓΘ, ΕΗ is equal to triangle ΑΒΓ. But that by ΓΘ, ΕΗ is a side of that from ΓΘ times that from ΕΗ. Therefore, the area of triangle ΑΒΓ times itself will become equal to that from ΘΓ times that from ΕΗ. (diagram 4) Let ΗΛ be drawn at right angles to ΓΗ, ΒΛ to ΓΒ. And let ΓΛ be joined. And so, since each of ΓΗΛ, ΓΒΛ is right, therefore, quadrilateral ΓΗΒΛ is in a circle (see theorem A below). Therefore, the angles enclosed by ΓΗΒ, ΓΛΒ are equal to two right angles (see theorem B below). (diagram 5) But those by ΓΗΒ, ΑΗΔ are also equal to two right angles since the angles at Η are bisected by ΑΗ, ΒΗ, ΓΗ and those by ΓΗΒ, ΑΗΔ are equal to those by ΑΗΓ, ΔΗΒ, and all of them are equal to four right angles (see theorem C below). (diagram 6) Therefore,the angle by ΑΗΔ is equal to the angle by ΓΛΒ. (diagram 7) But a right-angle, by ΑΔΗ, is also equal to a right-angle, by ΓΒΛ. Therefore, triangle ΑΗΔ is similar to triangle ΓΒΛ. As, therefore, ΒΓ is to ΒΛ, ΑΔ is to ΔΗ, that is ΒΘ to ΕΗ. And, alternando, as ΓΒ is to ΒΘ, ΒΛ is to ΕΗ (diagram 8) that is ΒΚ to ΚΕ, since ΒΛ is a parallel to ΕΗ. And, componendo, as ΓΘ is to ΒΘ, so is ΒΕ to ΕΚ. Thus, as that from ΓΘ is to that by ΓΘ, ΘΒ, so too is that by ΒΕΓ to that by ΓΕΚ, that is that from ΕΗ. For a perpendicular, ΕΗ, has been drawn in a rectangle from the right-angle to the base. Thus that from ΓΘ times that from ΕΗ, whose side was the area οf triangle ΑΒΓ, will be equal to that by ΓΘΒ times ΓΕΒ. And each of ΓΘ, ΘΒ, ΒΕ, ΓΕ is given. For ΓΘ is half of the perimeter of triangle ΑΒΓ, while ΒΘ is the excess by which half of the perimeter exceeds ΓΒ, and ΒΕ the excess by which half of the perimeter exceeds ΑΓ, and ΕΓ the excess by which half of the perimeter exceeds ΑΒ, since really ΕΓ is equal to ΓΖ, and ΒΘ to ΑΖ, since it is also equal to ΑΔ. Therefore, the area of triangle ΑΒΓ is also given.

In the following: **O**(x,y) is the rectangle formed by lines x and y, **T**(x) is the square on x, and **S(T**(x)) is the side of the square **T**(x)

Analysis: ΗΔ is perpendicular to ΑΒ, ΗΕ to ΒΓ, and ΗΖ to ΑΓ, since the sides of the triangle are tangent to the circle. So the radii are altitudes of triangles ΑΗΒ, ΒΗΓ, ΑΗΓ. Hence, triangle ΑΒΓ = ΒΗΓ + ΑΗΓ + ΑΗΒ = 1/2**O**(ΗΕ,ΒΓ) + 1/2**O**(ΔΗ,ΑΒ) +1/2**O**(ΗΖ,ΑΓ) = 1/2**O**(ΗΕ,ΑΒ+ΒΓ+ΓΑ) =**O**(ΗΕ, 1/2(ΑΒ+ΒΓ+ΓΑ)). So the difficulty is to find the radius. Since the two tangents from a point outside a circle are equal, ΔΒ = ΒΕ, ΕΓ= ΓΖ, and ΑΔ = ΑΖ. So ΑΒ + ΒΕ + ΕΓ = 1/2(ΑΒ+ΒΓ+ΓΑ). Since ΘΒ = ΑΒ, ΘΒ + ΒΕ + ΕΓ = 1/2(ΑΒ+ΒΓ+ΓΑ). Hence, **O**(ΗΕ,ΘΓ) = triangle ΑΒΓ.

triangle-ΑΒΓ · triangle-ΑΒΓ =** O**(ΗΕ,ΘΓ) · **O**(ΗΕ,ΘΓ) = **T**(ΗΕ) · **T**(ΘΓ). This is not a geometrical operation, but, notoriously, it is a good question what the operation is. In any case, it applies the penultimate step of Prop. 7.

Hero does not actually here draw the inference that triangle-ΑΒΓ =
$\sqrt{\mathrm{T({\rm H}{\rm E})\; \xb7\; T(\Theta \Gamma )}}$ = **O**(ΗΕ,ΘΓ)

The trick is to show that ΑΗΔ ≈ ΓΒΛ. Then, the rest is manipulations of proportions. By the similar triangles,

ΒΓ : ΒΛ = ΑΔ : ΔΗ = ΒΘ : ΕΗ (since. by stipulation, ΒΘ = Α Δ and radii ΕΗ = ΔΗ)

ΒΓ : ΒΘ = ΒΛ : ΕΗ = ΒΚ : ΚΕ (since triangles ΕΚΗ ≈ ΒΚΛ)

ΘΓ : ΒΘ = ΒΕ : ΚΕ (componendo: ΘΒ+ΒΓ : ΒΘ = ΒΚ+ΚΕ : ΚΕ)

**T**(ΘΓ) : **O**(ΒΘ,ΘΓ) = **O**(ΒΕ,ΕΓ) : **O**(ΓΕ,ΕΚ) (a : b = c : d ⇒ **O**(a,x) : **O**(b,x) = **O**(c,y) : **O**(d,y))

**T**(ΕΗ) = **O**(ΓΕ,ΕΚ) (becausse right triangles ΚΕΗ ≈ ΚΗΓ ≈ ΗΕΓ)

**T**(ΘΓ) : **O**(ΒΘ,ΘΓ) = **O**(ΒΕ,ΕΓ) : **T**(ΕΗ)

**T**(ΘΓ) · **T**(ΕΗ) =** O**(ΒΘ,ΘΓ) · **O**(ΒΕ,ΕΓ) (the meaning of this is difficult; see above)

triangle ΑΒΓ = $\sqrt{\mathrm{T({\rm H}{\rm E})\; \xb7\; T(\Theta \Gamma )}}$
= $\sqrt{\mathrm{O({\rm B}\Theta ,\Theta \Gamma )\; \xb7\; O({\rm B}{\rm E},{\rm E}\Gamma )}}$

ΒΘ = 1/2 perimeter - ΓΒ

ΘΓ = 1/2 perimeter

ΒΕ = 1/2 perimeter - ΑΓ

ΕΓ = 1/2 perimeter - ΑΒ

Theorem A: (diagram Thm A) If each diagonal of a quadrilateral makes a right angle with a side, the quadrilateral can be inscribed in a circle,

Proof: Let there be quadrilateral ABΓΔ and diagonals, ΑΓ, ΒΔ, with angles ΒΓΔ and ΑΒΓ righτ. Circumscribe triangle ΑΒΓ with circle ΑΒΓ (*El*. iv 5). I say that Δ is on the circumference of circle ΑΒΓΕ. Since angle
ΒΓΔ is right, ΑΔ is a diameter (*El*. iii 31). Since angle ΑΒΔ is also right, it too lies on a semicircle with diameter ΑΔ, i.e., the same semi-circle.

Theorem B: (diagram Thm B) The opposite angles together of a quadrilateral inscribed in a circle equal two right angles.

Let quadrilateral ΑΒΓΔ be inscribed in a circle, and diameters ΒΔ, ΑΓ be joined. I say that ΑΒΓ + ΓΔΑ = 2 right angles. Clearly, angles ΒΑΓ = ΒΔΓ, and ΔΒΓ = ΓΑΔ (*El*. iii 27). But also angles ΒΑΓ + ΓΑΔ + ΒΔΑ + ΑΒΔ = ΒΔΑ + ΔΑΒ + ΑΒΔ = 2 right angles *(*El. i 32). Hence, angles ΒΔΓ + ΔΒΓ + ΒΔΑ + ΑΒΔ = 2 right angles, but also ΒΔΓ + ΒΔΑ = ΑΔΓ, and ΔΒΓ + ΑΒΔ = ΑΒΓ. Hence, ΑΒΓ + ΑΔΓ = 2 right angles.

Similarly, we could show that ΒΑΔ + ΒΓΑ = 2 right angles.

return to demonstrion

Theorem C: If a quadrilateral has two adjacent sides equal, and opposite adjacent sides equal, then the line connecting the vertices of the two pairs of adjacent sides bisect the angles.

Let there be a quadrilateral ΑΒΓΔ with equal sides AB = BG and ΓΔ = ΔΑ. Let diameter ΒΔ be drawn. I say that ΒΔ bisects angle ΑΒΓ. Let ΓΑ be joined. Since, ΒΓ = ΓΔ, ΑΒ = ΑΔ, and ΑΓ is common, triangle ΑΒΓ = ADG. Therefore, angles ΒΑΓ = ΒΓΑ, and ΒΓΑ = ΔΓΑ.

It will be synthesized, in fact, in this way. Let ΑΒ be of 13 units, ΒΓ of 14 units, and ΑΓ of 15 units. Add 13 and 14 and 15. And they become 42. half of which become 21. Remove the 13. Remainders 8. Then the 14. Remainders 7. And further the 15. Remainders 6. 21 times 8, and what come-to-be times 7, and further what come-to-be times 6. They are combined 7056. The side of these of so much as 84 will be the area of the triangle.

9. And so, after we have learned how, the sides of a triangle being given, to find its area when the altitude is rational, let it be now to find the area when the altitude is not rational. Let there be a triangle ΑΒΓ, having ΑΒ of 8 units, ΒΓ of 10 units, and ΑΓ of 12 units. And let an altitude, ΑΔ, be drawn. Following, in fact, what was said in the case of the acute-angles, what’s twice by ΓΒΔ is of 20 units. Therefore ΒΔ will be of 1 unit, and that from it, therefore, of 1 unit. But also that from ΑΒ is of 64 units. Therefore a remainder, that from ΑΔ, will be of 63 units. But also that from ΒΓ is of 100 units. Therefore, that from ΒΓ times that from ΑΔ will be of 6300 units. A side of this is the [number] by ΒΓ, ΑΔ times itself; therefore the [number] by ΒΓ, ΑΔ times itself will be of 6300 units. Therefore, the half of that by ΒΓ, ΑΔ times itself will be of 1475. For where the sides of the squares are double one another, those from them are four times those from the halves. But the half of that by ΒΓ, ΑΔ is the area of the triangle. Therefore, the area of the triangle is in power 1475. But it is possible by taking the side of 63 units as near to find the area as if the altitude were rational. But the side 7 ∠ 4´ 8´ 16´ is near 63. And so it will be required by laying down the altitude of so much to find the area. It is 39 ∠ 8´ 16´.

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