Chapter VI: Book III: Concerning Petitions and Axioms (5)

But it is requisite to relate the other demonstrations of the present theorem, such as Heron, and the familiars of Porphyry have fabricated, without producing the right line, after the manner of Euclid. Let there be a triangle _a b c_, it is requisite, therefore, to shew, that the sides _a b_, _a c_, are greater than the side _b c_. Bisect the angle at _a_, by the right line _a e_. Because, therefore, the angle _a e c_, is external to the triangle _a b e_, it is greater than the angle _b a e_. But the angle _b a e_, was placed equal to the angle _e a c_. The angle, therefore, _a e c_, is greater than the angle _e a c_. Hence, the side also _a c_, is greater than the side _c e_. And for the same reason the side _a b_, is greater than the side _b e_. For the angle _a e b_ is external to the triangle _a e c_, and is greater than the angle _c a e_; that is than the angle _e a b_. And on this account the side _a b_, is greater than the side _b e_. The sides, therefore, _a b_, _a c_, are greater than the whole side _b c_. And the like may be shewn of the other sides. Let there again be a triangle _a b c_. If therefore the triangle _a b c_, be equilateral, two sides will be doubtless greater than the remaining one: for when there are three equal quantities, any two are double of the remainder. But if it be isosceles, it will have a base either less, or greater than each of the equal sides. If therefore the base be less, the two sides are given greater than the remainder. But if the base be greater, let it be _b c_, and cut off from it a part equal to either of the sides, which let be _b e_, and connect _a e_. Because, therefore, the angle _a e c_, is external to the triangle _a e b_, it is greater than the angle _b a e_. On the same account the angle _a e b_, is greater than the angle _c a e_. Hence, the angles about the point _e_, are greater than the whole angle about the point _a_, of which _b e a_ is equal to _b a e_, since _a b_ is equal to _b e_. The remainder, therefore, _a e c_, is greater than the remainder _c a e_. Hence, the side _a c_, is greater than the side _c e_. But the side _a b_, was also equal to the side _b e_. The sides, therefore, _a b_, _a c_, are greater than the side _b c_.

But if the triangle _a b c_, be scalene, let the greatest side be _a b_, the middle _a c_, and the least _c b_. The greatest side, therefore, assumed with either of the others, exceeds the remainder: for by itself it is greater than either. But if we are desirous of shewing that the sides _a c_, _c b_, are greater than the greatest side _a b_, we must employ the same construction as in the isosceles triangle, cutting off from the greater side, a part equal to one of the other sides, and connecting the line _c e_, and using the external angles of the triangles.

Let there be again any triangle _a b c_. I say that the sides _a b_, _a c_, are greater than the side _b c_. For if they are not greater, they are either equal or less. Let them be equal, and cut off _b e_, equal to _a b_. The remainder, therefore, _e c_, is equal to _a c_. Because then, _a b_, _b e_, are equal, they subtend equal angles; and this is likewise true of _a c_, _e c_, because they are equal. Hence, the angles at the point _e_, are equal to the angles at the point _a_, which is impossible. Again let the sides _a b_, _a c_, be less than _b c_, and cut off _b d_, equal to _a b_, and _e c_ to _a c_. Because, therefore, _a b_ is equal to _b d_, the angle _b d a_, is not unequal to the angle _b a d_. And because _a c_ is equal to _c e_, the angle _c e a_, is equal to the angle _e a c_. Hence, the two angles _b d a_, _c e a_, are equal to the two _b a d_, and _e a c_. Again, because the angle _b d a_ is external to the triangle _a d c_, it is greater than the angle _e a c_: for it is greater than _c a d_. By a similar reason also, because the angle _c e a_, is external to the triangle _a b e_, it is greater than the angle _b a d_: for it is greater than the angle _b a e_. Hence, the angles _b d a_, _c e a_, are greater than the two _b a d_, _e a c_. But they were also equal to them, which is impossible. The sides, therefore, _a b_, _a c_, are neither equal to, nor less than the side _b c_, but greater. And the like may be exhibited in others.

PROPOSITION XXIV. THEOREM XIV.

If upon one side of a triangle, two right lines beginning from
the extremities, are internally constituted, the constituted
right lines will be less than the other sides of the triangle,
but they will contain a greater angle.

That which is expressed by the proposition, is, indeed, manifest; and the demonstration adopted by the elementary institutor, is evident; and the theorem is consequent to the first principles, since it depends on two theorems, the one previously exhibited, and the sixteenth. For in order to shew, that the lines internally constituted, are less than the external, the theorem is required, which says, _the two sides of every triangle, are greater than the remaining one_: but for the purpose of confirming that the angle comprehended by them is greater than that comprehended by the external sides, that theorem procures the greatest utility, which says, _the external angle of every triangle, is greater than the internal and opposite angle_. But you will receive at the same time, conviction of geometrical diligence, and a commemoration of things admirable in the mathematical disciplines, if we shall shew that _it is possible within a certain triangle, upon one of its sides, not upon the whole, but upon some one of its parts, to constitute two right lines greater than the external right lines[22]; and again, others comprehending a less angle, and comprehended in the angle made by the external lines_. For this being exhibited, it will at the same time be manifest, that the institutor of the Elements necessarily adds, that the internally constituted lines must begin from the extremities of the common basis; and must be constituted upon one whole side, and not upon any one of its parts: but likewise, as I have said, one of the admirable things which geometry contains, will be manifest. For is it not, indeed, admirable, that the lines constituted upon the whole side, should be less than the external sides: but that those constituted upon a part should be greater? Let there be then a right angled triangle _a b c_, having the angle at the point _b_ right, and take in the side _b c_, any point _d_, and connect _a d_. Hence, _a d_ is greater than _a b_. Take from _a d_, a part equal to _a b_, which let be _d e_, and bisect _e a_, in the point _f_, and connect _f c_. Because, therefore, _a f c_ is a triangle, the lines _a f_, _f c_, are greater than _a c_. But _a f_ is equal to _f e_. The right lines therefore, _f e_, _f c_, are greater than _a c_. But _d e_ is equal to _a b_. Hence, the right lines _f c_, _f d_, are greater than the right lines _a b_, _a c_, and they are internally constituted.

Let there be again an isosceles triangle _a b c_, having the base _b c_, greater than either of the equal sides. Then from _b c_, cut off _b d_, equal to _a b_, and connect _a d_, and take in _a d_, any point _e_, and connect _e c_. Because, therefore, _a b_, is equal to _b d_, the angle _b a d_, is also equal to the angle _b d a_. And because the angle _b d a_ is external to the triangle _e d c_, it is greater than the internal and opposite _d e c_. Hence, the angle _b a d_, is greater than the angle _d e c_. Much more, therefore, is the angle _b a c_, greater than the angle _d e c_; and _b a c_ is contained by the external lines, but _d e c_ by internal lines. Within a triangle, therefore, right lines _d e_, _e c_, comprehending a lesser angle, are constituted within the angle comprehended by the external lines; and the thing proposed is shewn without employing the parallel lines of expositors. Hence, it is necessary that the constituted right lines should begin from the extremities of the basis: for those which are constituted upon any one of its parts, are shewn to be sometimes greater than the external lines, and to comprehend a lesser angle. But when they are constituted in this manner, beginning from the extremities, the species of triangles, called (ἀκιδοειδῆ) or, similar to the point of a spear[23], presents itself to our view; and is one of the admirable things contained in geometry, viz. to find a quadrilateral triangle. As for example, the triangle A B C. For it is contained by the four sides B A, A C, B D, D C; but it has three angles, one at B, the other at A, and the other at C. And hence, the present figure is a quadrilateral triangle.

PROPOSITION XXII. PROBLEM VIII.

To construct a triangle from three right lines, which are equal
to three given right lines. But it is requisite that two of the
lines must be greater than the remaining one, in whatever manner
they may be taken.

We again pass to problems, and Euclid commands us to construct a triangle from three proposed right lines, two of which are greater than the remaining one, equal to given right lines. Because he knew this in the first place, that it was impossible to construct a triangle from those same lines, which had already received the declared position: but that this was possible to be effected from their equals. In the next place, he knew it was necessary that two of the right lines about to complete the triangle, should be greater than the remaining one: for the two sides of every triangle are greater than the remaining one, however assumed, as we have shewn. On this account he adds, _that it is necessary the first right lines remaining, to construct a triangle from three equal to them: but that it is requisite, any two, however taken, should be greater than the remainder, or there will not be a triangle from three lines equal to the given right lines_. But by this means he also destroys all the objections which are urged against the construction, and which may be perfectly dissolved by this addition. Hence, the present problem ranks among things determined, and not among such as are indetermined: For of problems as well as of theorems, some are indeterminate, but others without termination. Thus if we should simply say, _from three right lines which are equal to three given right lines, to construct a triangle_, the problem is indeterminate and impossible. But if we add, _two of which, however assumed, are greater than the remainder_, the problem is determined and possible. For as the division of theorems takes place, according to true and false, so that of problems according to a possible and impossible enunciation. But that the objections which are urged against the construction, may be from hence dissolved, we shall learn from a little inspection: for we shall follow the words of the geometrician. Let there be three right lines, _a b c_, of which any two, however taken, are greater than the third, and let it be required to accomplish the thing proposed. Let there be placed a certain right line _d e_, on one part finite, as at the point _d_: and on the other part infinite. Then place _d f_ equal to _a_, but _f g_ to _b_: and _g h_ to _c_. And from the centre _f_, but interval _f d_, let a circle _k_ be described. Again, with the centre _g_, but interval _g h_, let the circle _l_ be designed; and the circles will intersect each other. For this is assumed by the institutor of the Elements. But it may be asked how this takes place? For perhaps they either only touch each other, or they do not even touch. Since it is necessary that they should suffer some one of three cases, I mean that they should either intersect or touch, or be distant from each other. I say, therefore, that they necessarily intersect each other. For let them in the first place, touch each other. Because, therefore, the point _f_ is the centre of the circle _k_, _d f_ is equal to _f n_. And because the point _g_ is the centre of the circle _l_, _h g_ is equal to _g m_. The two, therefore, _d f_, _g h_, are equal to one, viz. to _f g_. But they were placed greater than one, as also _a_, together with _c_, is greater than _b_. They are therefore equal to it, and at the same time greater, which is impossible. Again, if it be possible, let the circles be distant from each other, as _k_ and _l_. Because, therefore, the point _f_, is the centre of the circle _k_, _d f_, is equal to _f n_. And because the point _g_, is the centre of the circle _l_, _h g_ is equal to _g m_. The whole, therefore, _f g_, is greater than the two, _d f_, _h g_: for _f g_, exceeds _d f_, _g h_, by _n m_. But it was supposed that _d f_, _h g_, were greater than _f g_, in the same manner as _a_ and _c_ are greater than _b_. For _d f_ was placed equal to _a_, but _f g_, to _b_, and _h g_ to _c_. It is necessary, therefore, that the circles _k_ _l_, should intersect each other. Hence, the institutor of the Elements very properly receives them cutting one another: since of the three right lines, he supposes two greater than the third, however, they may be assumed, but neither equal to, nor less than one. But it is necessary that when the circles touch, two of the lines should be equal to the third; and that when they are distant from each other, two should be greater than the remainder.

PROPOSITION XXIII. PROBLEM IX.

On a given right line, and at a given point in it, to constitute an
angle equal to a given right lined angle.

This also is a problem, whose invention according to Eudemus is rather the gain of Oenopides than of Euclid: but it requires the construction of an angle, on a given right line; and at a given point in it equal to another right lined angle. This, then, Euclid necessarily adds, that the given angle must be rectilineal; because it is impossible that an angle can be construed on a right line equal to every angle. For it has been shewn[24] that there are only two curve-lined angles equal to right-lined angles, viz. the angle of a lunular figure, which we have proved equal to every right-lined angle; and the angle of that figure similar to an axe[25], which is equal to two thirds of a right angle. But a lunular figure of this kind, which is called (πελεκοειδὲς) Pelecoides, is formed from two circles cutting each other through their centres. However, the construction of an angle on a certain right line, causes the constituted angle to become determinate, and not indifferent in species, but forms it either right-lined, or mixt. But since no mixt can be equal to a right-lined angle, it is manifest that this must be perfectly rectilineal. The institutor of the Elements, therefore, simply using the present problem, and constructing a triangle from three right lines, equal to three given lines, accomplishes the thing proposed. But you may receive a more exquisite construction of the triangle, by the following method. Let there be a given right line _a b_, and a given point in it _a_, and a given right-lined angle _c d e_. It is required, therefore, to accomplish the problem. Connect _c e_, and produce _a b_ on both sides to the points _f_, _g_. Then place _f a_, equal to _c d_, and _d e_ to _a b_, and _b g_ to _e c_. And with the centre _a_, but interval _a f_, describe the circle _k_. And again, as in the preceding, with the centre _b_, but interval _b g_, describe the circle _l_. The circles, therefore, will cut each other, as we have shewn in the last proposition. Let them cut each other in the points _m_, _n_, and from these points draw right lines to the centres as in the figure. Because, therefore, _f a_, is equal to _a m_, and _a n_, but _c d_, is equal to _f a_; _a m_, and _a n_, will be each equal to _c d_. Again, because _b g_ is equal to _b m_, and _b n_, but _g b_ is not unequal to _c e_; _b m_, and _b n_, will be also equal to _c e_. But _a b_ is equal to _d e_. The two therefore, _a b_, _a m_, are not unequal to the two _d e_, _d c_, and the base _b m_, is equal to the base _c e_. Hence, the angle _m a b_, is equal to the angle at the point _d_. And again, the two _n a_, _a b_, are equal to the two _c d_, _d e_, and the base _n b_, is equal to the base _c e_. The angle, therefore, _n a b_, is equal to the angle _c d e_, and the thing proposed is doubly accomplished: for we have not only constituted one, but two angles, equal to the given angle, on each side of the right line _a b_; so that in whatever part we may desire the construction to be made, it will be indubitable, and without contradiction. And this we have added to the construction of the elementary institutor.

But we cannot praise the method of Apollonius, because it requires the assistance of the third book. For he receives any angle _c d e_, and a right line _a b_, and with the centre _d_, but interval _c d_, he describes a circumference _c e_. In like manner with the centre _a_, but interval _a b_, he describes a circumference _b f_; and intercepting a circumference _c e_, equal to _b f_, he connects the right line _a f_, and affirms that the angles _a_ and _d_, insisting on equal circumferences, are equal. But it is necessary to pre-assume that _a b_ is equal to _c d_, in order that the circles may be also equal. We therefore think that a demonstration of this kind requiring posterior propositions, is foreign from an elementary institution; and we give the preference to that of the geometrician, as consequent to principles.

PROPOSITION XXIV. THEOREM XV.

If two triangles have two sides equal to two, each to each, but
the one angle contained by the equal right lines greater than
the other: they shall also have the base of the one greater than
the base of the other.

Euclid again passes on to theorems, and speaks concerning inequality in two triangles, in a manner similar to his discourse concerning equality. For supposing two triangles, having two sides equal to two, each to each, he sometimes places the vertical angle equal in each, and sometimes unequal; and he proceeds in a similar manner with respect to the base. Besides this, he demonstrates that the equality of the bases is consequent to the equality of the vertical angle, and that the equality of the vertical angles, is consequent to the equality of the bases: but he now shews that the inequality of the one, follows the inequality of the other. The present theorem, therefore, is opposite to the fourth: for that, indeed, supposes the vertical angles of the triangles equal, but this supposes them unequal. And that demonstrates the equality of their bases; but this proves them unequal, in the same manner as their angles. It precedes, however, the following theorem: for that deduces its proof of inequality from the bases to the angles subtending the bases: but this, on the contrary, reasons from the angles to the bases, which are under the angles. Hence it is, after this manner, the converse of its consequent proposition, but opposite to the eighth theorem. For the one from the equality of the bases, demonstrates the equality of the vertical angles, but the other from the inequality of the bases, shews that the vertical angles are unequal. It is, however, common to these four (two of which are conversant with equality, I mean the fourth, and the eighth, but two about inequality, the present and the following; and two begin from angles, viz. the fourth, and the object of investigation in the present, but two from bases, viz. the eighth, and the following proposition); it is common, I say, to all these four, as well to the fourth and the eighth, as to the twenty-fourth and twenty-fifth, to have two sides equal to two, each to each. For these being unequal, all enquiry is superfluous, and subject to deception. And thus much for a universal speculation concerning the present theorem.

But let us now consider the construction of the elementary institutor, and add to it where deficient. For Euclid receiving two triangles, _a b c_, _d e f_, having the sides _a b_, _a c_, equal to the sides _d e_, _d f_, each to each, and the angle at the point _a_, being greater than the angle at the point _d_, and willing to shew that the base _b c_, is greater than the base _e f_, on the right line _e d_, and at a point in it _d_, constitutes an angle _e d h_, equal to the angle at the point _a_. For the angle at the point _a_, is greater than the angle at the point _d_, and he connects _d h_, equal to _a c_. The right line, therefore, _e h_, produced to the point _h_, either falls above, or upon, or beneath the line _e f_. The institutor of the Elements, indeed, considers it as lying above the line. But let it be upon the right line. Again, therefore, we may exhibit the same. For the two _a b_, _a c_, are equal to the two _d e_, _d h_, and they contain equal angles. Hence, the base _b c_, is equal to the base _e h_. But _e h_ is greater than _e f_; and on this account _b c_ is greater than _e f_. Again, let it be placed beneath _e f_. Connecting, therefore, _e h_, we must say, that since _a b_, _a c_, are equal to _d e_, _d h_, and they comprehend equal angles, _b c_ is also equal to _e h_. Because, therefore, within the triangle _d e h_, two right lines _d f_, _f e_, are constructed on the side _d e_, they are less than the external sides. But _d h_, is equal to _d f_: for it is equal to _a c_. Hence _h e_ is greater than _e f_. But _h e_ is equal to _b c_. And therefore, _b c_ is greater than _e f_. The theorem, therefore, is exhibited according to every position. Why then, as is the fourth theorem, he at the same time demonstrated that the areas of triangles are equal, does he not add in the present, that besides the inequality of the bases, the areas also are unequal? Against this doubt we must say, that there is not the same proportion in equal, as in unequal angles and bases. For when the angles and bases are equal, the equality also of the triangles follows: but when they are unequal, it is not necessary that the inequality of the areas should be consequent; since the triangles may as well be equal, as unequal; and that may be greater, and likewise less, which contains the greater angle, and the greater base. On this account, therefore, the institutor of the Elements leaves the comparison of the triangles; to which we may add, that the contemplation of these, requires the doctrine of parallels.

But if it be requisite, that anticipating things which are afterwards exhibited, we at present make a comparison of areas, we must say, that if the angles _a_, _d_, are equal to two right, the triangles may be shewn to be equal: but when they are greater than two right, the lesser triangle will be that which contains the greater angle; and when they are less than two right, this will be the case with the greater triangle. For let the construction in the element be given, and produce _e d_, _f d_, to the points _k_, _h_; and let us suppose the angles _b a c_, _e d f_, equal to two right. Because, therefore, the angle _b a c_, is equal to the angle _e d g_, the angles _e d g_, _e d f_, are equal to two right. But the angles _e d g_, _k d g_, are also equal to two right. Let the common angle _e d g_ be taken away, and the remainder _e d f_, will be equal to the remainder _k d g_. But _e d f_ is equal to _h d k_; for they are vertical angles. Hence, the angle _k d g_, is equal to the angle _h d k_. And because the angle _g d h_, is external to the triangle _g d f_, it is equal to the two internal and opposite angles at the points _g_ and _f_. But these angles are equal to each other, because _d g_ is equal to _d f_. Hence, the angle _g d h_, is double of the angle at the point _g_, and of the angle at the point _f_. The angle, therefore, at the point _g_, is equal to the angle _g d k_, and they are alternate; and consequently _d e_ is parallel to _f g_. The triangles, therefore, _g d e_, _f d e_, are upon the same base _d e_, and between the same parallels _d e_, _g f_; and are consequently equal. But the triangle _g d e_, is equal to the triangle _a b c_; and so the triangle _d e f_, is not unequal to the triangle _a b c_. And here you may observe, that we require three theorems belonging to the doctrine of parallels; one, indeed, affirming, _that the external angle of every triangle is equal to the two internal and opposite angles_: but the other, _that if a right line falling upon two right lines, makes the alternate angles equal, the right lines are parallel_; and the third, _that triangles constituted upon the same base, and between the same parallels, are equal_, which the institutor of the Elements also knowing, omits the comparison of triangles.

But let the angles _b a c_, _e d f_, be greater than two right, and let the same things be constructed. Because, therefore, the angles _b a c_, _e d f_, i.e. the angles _e d g_, _e d f_, are greater than two right; but the angles _e d g_, _g d k_, are equal to two right, by taking away the common angle _e d g_, the angle _e d f_, is greater than the angle _g d k_. Hence, the angle _g d h_, is more than double of the angle _g d k_; and so the angle _g d k_, is less than the angle at the point _g_. Let _g d k_ be placed equal to _d g l_, and let _e l_, and _d l_, be connected: _g l_, therefore, is parallel to _d e_; and hence, the triangles _g d e_, _l d e_, are equal. But the triangle _l d e_, is less than the triangle _f d e_. The triangle, therefore, _g d e_, is less than the triangle _f d e_. But the triangle _g d e_, is equal to the triangle _a b c_; and hence, the triangle _a b c_, is less than the triangle _f d e_, viz. is less than the triangle which contains the greater angle.

In the third place, let the unequal angles be less than two right, and let the same things be constructed. Because, therefore, the angles _e d g_, _g d k_, are equal to two right, by taking away the common angle _e d g_, the whole _g d h_, is less than double of _g d k_. But it is double also of the angle at the point _g_. Hence, the angle _g d k_, is greater than the angle at the point _g_. Let the angle _d g l_, be placed equal to the angle _g d k_, and let _g l_ coincide with _e l_, in the point _l_, and connect _d l_. Hence, _g l_ is parallel to _d e_; and consequently the triangles _g d e_, _l d e_, are equal to each other. But the triangle _l d e_, is greater than the triangle _f d e_; and the triangle _g d e_, is equal to the triangle _a b c_. Hence, the triangle _a b c_, is greater than the triangle _d f e_. It is shewn, therefore, that the triangle _a b c_, is both equal to, and is also greater and less than the triangle _d e f_, the angles at the points _a_ and _d_, being either equal to, or greater or less than two right. And thus, all the hypotheses may be accomplished. For what if the angle at the point _a_, should be one right, and the half of a right angle, but the angle at the point _b_, the half of one right, would not those two angles be equal to two right? But what if the angle at the point _a_, should be one right, and the half of a right, but the angle at the point _b_, two thirds of one right, would they not be greater than two right angles? And lastly, if the angle at the point _a_, should be one right, and the half of a right angle, but the angle at the point _b_, a third part of a right angle, would they not be less than two right, and the angle _a_ be greater than the angle _d_? All these comparisons, therefore, are produced by the assistance of parallels; and hence, they are necessarily not found in the present elementary institution.

PROPOSITION XXV. THEOREM XVI.

If two triangles have two sides equal to two, each to each, but
have the base of the one greater than the base of the other;
they shall likewise have the angle contained by the equal sides
in the one, greater than the angle contained by the equal sides
in the other.

The present theorem is the opposite to the eighth, but the converse of the preceding. For the institutor of the Elements produces theorems concerning the equality and inequality of angles and bases, according to _conjunction_; in each of the _conjunctions_, receiving some as precedents, but others as converse. And in such as are precedent indeed, he employs direct ostensions: but in such as are converse, he uses deductions to an impossibility. After this manner he proceeds in some particular triangle, sometimes from the equality of the sides which it contains, shewing the consequent equality of the angles which they subtend: but sometimes from their inequality evincing inequality. And again, on the contrary, affirming that equality of sides is consequent to equality of angles, but inequality to inequality. However, that we may proceed to the thing proposed, we refer those who are desirous of learning how the geometrician shews when this is manifest, to his books on this subject. But we shall briefly relate the demonstrations which others produce of this proposition; and in the first place, that which Menelaus Alexandrinus invented and delivered. Let there be two triangles _a b c_, _d e f_, having the two sides _a b_, _a c_, equal to the two _d e_, _d f_, each to each, and the base _b c_, greater than the base _e f_, I say that the angle at the point _a_, is greater than the angle at the point _d_. For let there be cut from the base _b c_, a line _b g_, equal to the base _e f_, and construct at the point _b_, an angle _g b h_, equal to the angle _d e f_, and place _b h_ equal to _d e_. Lastly, connect _h g_, and produce it to the point _k_, and connect _a h_. Because, therefore, _b g_ is equal to _e f_, but _b h_ to _e d_, the two are equal to the two, and they contain equal angles. Hence, _g h_ is equal to _d f_, and the angle _b h g_, is not unequal to the angle _e d f_. And because _g h_ is equal to _d f_, but _d f_ to _a c_, _g h_, also, is equal to _a c_. Hence _h k_ is greater than _a c_, and consequently is much greater than _a k_. The angle, therefore, _k a h_, is greater than the angle _k h a_. Again, because _b h_, is equal to _a b_, for it is equal to _d e_, the angle _b h a_, is equal to the angle _b a h_. Hence, the whole angle _b h k_, is less than the whole, _b a c_, but is shewn to be equal to the angle at the point _d_. The angle, therefore, _b a c_, is greater than the angle at the point _d_. And such is the demonstration of Menelaus.

But Heron, the mechanist, shews the same thing, in the following manner, without leading to an impossibility, as is the case with the demonstration of Euclid. Let there be two triangles _a b c_, _d e f_, with the same hypotheses as above. And because _b c_ is greater than _e f_, let _e f_ be produced, and place _e g_ equal to _b c_; and in like manner extend _d e_, and place _d h_ equal to _d f_. The circle, therefore, which is described with the centre _d_, and interval _d f_, will pass also through the point _h_. Let it be described as _f k h_. And because _a c_, _a b_, are together greater than _b c_, but these are equal to _e h_, and _b c_ is equal to _g e_, hence the circle which is described with the centre _e_, but interval _e g_, will cut _e h_. Let it cut _e h_, as the circle _g k_, and connect from the common section of the circles to the centres, the right lines _k d_, _k e_. Because, therefore, the point _d_, is the centre of the circle _h k f_, _d k_ is equal to _d h_, i.e. to _a c_. Again, because the point _e_, is the centre of the circle _g k_, the line _e k_, is equal to _e g_, i.e. to _b c_. Hence, since the two _a b_, _a c_, are equal to the two _d e_, _d k_, and the base _b c_, is equal to the base _e k_, the angle, also, _b a c_, is equal to the angle _e d k_. And thus the angle _b a c_, is greater than the angle _f d e_.

PROPOSITION XV. THEOREM XVII.

If two triangles have two angles equal to two, each to each,
and one side equal to one side, either that which is adjacent
to the equal angles, or that which subtends one of the equal
angles: then they shall have the remaining sides equal to the
remaining sides, each to each, and the remaining angle equal to
the remaining angle.

It is necessary, that he who wishes to compare triangles with each other, according to sides, angles, and areas, should either, by receiving the sides alone equal, enquire after the equality of angles; or by assuming the angles alone equal, investigate the equality of the sides; or by mingling the angles and sides, scrutinize the equality of angles and sides. Since, therefore, Euclid alone receives the angles equal, he could not likewise shew that the sides of the triangles are equal. For the least triangles are equiangular with the greatest, though at the same time they are excelled by them, both according to sides and comprehended space: but the angles of the former are separately equal to the angles of the latter. However, as he supposes the sides alone to be equal, he demonstrates that all are equal, by the eighth theorem, in which there are two triangles having two sides equal to two, each to each, and the base to the base, and these are shewn to be equiangular, and to possess a power of comprehending equal spaces. And the institutor of the Elements omits this addition, as necessarily following from the fourth, and requiring no demonstration. But when receiving sides and angles, he ought to receive either one side equal to one, and one angle equal to one; or one side, and two angles of the triangles, equal to two; or on the contrary, one angle and two sides; or one angle and three sides; or one side and three angles; or more than one side, and more than one angle. But when he had received one angle, and one side, he could by no means shew the thing proposed. I mean, the equality of the rest. For it is possible that two triangles which are equal, according to one side only, and one angle, may be entirely unequal as to the rest. Thus let there be a right line _a b_, perpendicularly erected upon the right line _c d_, but let _b d_ be greater than _b c_, and connect _a c_, _a d_. In these triangles, therefore, there is one common side, and one angle equal to one, but all the rest are unequal. But it is lawful to receive one side, and two angles, and to prove the rest equal, and this he performs by the present theorem: though again, to suppose one side, and three equal angles, is superfluous; since from the equality of two alone, the equality of the rest is exhibited. Again, receiving one angle, and two equal sides, he demonstrates that the rest are equal in the fourth theorem. But it is superfluous to receive one angle, and three equal sides: for two equals being alone assumed, conclude the equality of the rest. Besides, it is superfluous to assume two sides, and two equal angles; or two sides, and three equal angles; or two angles and three sides; or three angles and three sides. For the consequents to fewer hypotheses attend likewise a greater multitude, while the hypotheses are received with proper conditions. Hence, three hypotheses requiring demonstration, present themselves to our view, one, which alone receives three sides; and another which assumes one side, and two angles, which the geometrician now proposes; and a third, the opposite to this. On this account, we have only these three theorems, concerning the equality of triangles, which are conversant in sides and angles; since all the other hypotheses are either invalid for the purpose of shewing the object of enquiry; or they are valid indeed, but superfluous, because the same things may be readily procured by fewer hypotheses. As, therefore, when he assumed two sides equal to two, and one angle equal to one, he did not, indeed, assume every angle, but (as it was proposed by him) that contained by equal right lines, in the same manner when he assumes two angles equal to two, and one side to one, he does not assume any side, but either that which is adjacent to the equal angles, or that which subtends one of the equal angles. For neither is it possible in the fourth theorem, by assuming any equal angle, nor in the present by assuming any side, to shew the equality of the rest.

Thus for example, an equilateral triangle _a b c_, being given, let the side _b c_ be divided into unequal parts, by the line _a d_. Hence, there will be formed two triangles, having two sides _a b_, _a d_, equal to the two _a c_, _a d_, and one angle at the point _b_, equal to one angle at the point _c_, but the remaining sides will not also be equal, as for instance, the side _b d_, to the side _d c_: for they are unequal. But neither are the remaining angles equal: the reason of which is, because we receive an angle equal to an angle, but not the angle which is contained by equal sides. After the same manner, indeed, the present theorem also will appear dubious, unless we assume, according to the aforesaid condition, an equal side subtending one of the equal angles, or adjacent to the equal angles. For let there be a right angled triangle _a b c_, having the angle at the point _b_ right, and the side _b c_, greater than the side _b a_, and let there be constructed on the right line _b c_, and at a point in it _c_, an angle _b c d_, equal to the angle _b a c_, and let _b d_, _c d_, produced, coincide in the point _d_. There are two triangles, therefore _a b c_, _b c d_, having one side _b c_ common, and two angles equal to two, viz. _a b c_, to _c b d_ (for they are right), and _b a c_ to _b c d_, according to construction. Hence, as it appears the triangles are equal, and yet it may be shewn that the triangle _b d c_, is greater than the triangle _a b c_. But the reason of this is, because in the triangle _a b c_, we assume the common side _b c_, subtending one of the equal angles, viz. the angle at the point _a_: but in the triangle _b c d_, we assume the equal side, adjacent to the equal angles. It was requisite, therefore, in each, either to subtend one of the equal angles, or to be adjacent to the equal angles. But not observing this, we affirmed that triangle to be equal, which is necessarily greater: for is not the triangle _b c d_, greater than the triangle _a b c_? To be convinced of this, let there be constructed on the right line _b c_, and at a given point in it _c_, an angle _f c b_, equal to the angle _a c b_: for the angle _b c d_, as well as the angle at the point _a_, is greater than the angle _a c b_. Because, therefore, there are two triangles, _a b c_, _b c f_, having two angles _a b c_, _b c a_, equal to two _c b f_, _b c f_, each to each, and one side common, adjacent to the equal angles, viz. _b c_, the triangles are equal. But the triangle _b c d_ is greater than the triangle _b c f_, and consequently it is also greater than the triangle _a b c_. But it was formerly shewn to be equal, on account of the assumption of any side: And thus much the diligence of Porphyry has supplied us on the present occasion. But Eudemus, in his Geometrical Narrations, refers the present theorem to Thales. For he says it is necessary to use this theorem in determining the distance of ships at sea, according to the method employed by Thales in this investigation. But from the preceding division we may briefly assume all the contemplation concerning the equality of triangles, and are enabled to relate the causes of things omitted, confuting those hypotheses, as either false, or superfluous. And thus far we determine the limits of the first section of the elementary institutor, because he forms the constructions and comparisons of triangles, according to equal and unequal. And by construction, indeed, he delivers their essence: but by comparison, their identity and diversity. For there are three things which are conversant about being, _essence_, _same_, and _different_[26], as well in quantities, as in qualities, according to the propriety of subjects. From these, therefore, as images it may be shewn, that every thing is the _same_ with itself, and _differs_ from itself, on account of the multitude which it contains; and that all things are the _same_ with one another, and _different_ from themselves. For both, in every triangle, and in more triangles than one, equality and inequality has been found to reside.