D G greater than the other, and at the point D, in the straight line DE, make (I. 23.) on the same side with EDF the angle EDG equal to BAC; make also DG equal (I. 3.) to AC or DF, and join EG, GF. * Because AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two ED, DG, each to each, and the angle BAC is equal to the angle EDG; therefore the base BC is equal (I. 4.) to the base EG. Again, because DG is equal to DF, the angles (I. 5.) DFG, DGF are equal: but the angle DGF is greater (I. ax. 9.) than EGF; therefore the angle DFG is greater than EGF; and much more is the angle EFG greater than the angle EGF. And because the angle EFG of the triangle EFG is greater than its angle EGF; the side EG is greater (I. 19.) than the side EF; but EG is equal to BC; and therefore also BC is greater than EF. Therefore, if two triangles, &c. E B F IF two triangles have two sides of the one equal to two sides of the other, each to each, but their bases unequal; the angle contained by the sides of that which has the greater base, is greater than the angle contained by the sides equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two DE, DF, each to each ; but the base BC greater than the base EF: the angle A is likewise greater than the angle D. For, if it be not greater, it must either be equal to it, or less; but the angle A is not equal to D, because then (I. 4.) the base BC would be equal to EF; but it is not. Neither is it less; because then (I. 24.) the base BC would be less than EF; but it is not; therefore the angle A is not less than the angle D; and it B с E F * Hence the angle DGE is not greater (I. 5. and 18.) than DEG ; but DHG is greater (I. 16.) than DEG: therefore DHG is greater than DGH, and (I. 19.) DG or DF is greater than DH ; and consequently the point H lies between D and F, and the line EG above EF. If, on the contrary, a line equal to DE the less side, were drawn through D, making with DF, on the same side of it with DE, an angle equal to A, the extremity of that line might fall on FE produced, or above it or below it; and thus the proof would require three cases, while the method here given, which is that of Simson, is universally applicable. Let the student compare this proposition, and the following with the fourth and eighth of this book. has been shown, that it is not equal to it: therefore, the angle A is greater than D. Wherefore, if two triangles, &c. PROP. XXVI. THEOR. D F If two angles of one triangle be equal to two angles of another, each to each, and if a side of the one be equal to a side of the other similarly situated in respect to those angles ; then (1.) the remaining sides are equal, each to each ; (2.) the remaining angles are equal; and (3.) the triangles are equal to one another. Let ABC, DEF be two triangles which have the angles ABC, BCA equal to the angles DEF, EFD, each to each, viz., ABC to DEF, and BCA to EFD; also one side equal to one side: and, G First, let those sides be equal to which the angles are adjacent, that are equal in the two B triangles ; viz., BC to EF: the other sides are equal, each to each, viz., AB to DE, and AC to DF; and the third angle BAC to the third angle EDF; and the areas of the triangles are equal. For, if AB be not equal to DE, one of them must be the greater. Let AB be the greater, and make (I. 3.) BG equal to DE, and join GC. Then, because GB is equal to DE, and (hyp.) BC to EF, the two sides GB, BC are equal to the two DE, EF, each to each ; and (hyp.) the angle GBC is equal to the angle DEF; therefore, the angle GCB is equal (I. 4. part 3.) to DFE; but DFE is (hyp.) equal to ACB; wherefore, also the angle GCB is equal to ACB, a part to the whole, which (I. ax. 9.) is impossible; therefore AB is not unequal to DE, that is, it is equal to it. Then, since BC is equal to EF, the two AB, BC are equal to the two DE, EF, each to each ; and the angle ABC to DEF: therefore (1. 4.) the base AC is equal to the base DF, the triangle ABC to the triangle DEF, and the third angle BAC to the third angle EDF. Next, let the sides which are opposite to equal angles in each triangle be equal to one another, viz., AB to DE; likewise in this case, the other sides are equal, viz., AC to DF, and BC to EF; also the triangle ABC to DEF, and the third angle BAC to the third angle EDF. For, if BC be not equal to EF, let BC be the greater, and make BH equal to EF, and join AH. Then, because BH is equal to EF, and AB to DE; the two AB, BH are equal to the two DE, EF, each to each ; and they contain equal angles; therefore (1. 4. part 3.) the angle BHA is equal to EFD: A D B нс E but EFD is equal (hyp.) to BCA; therefore also the angle BHA is equal to BCA, that is, the exterior angle BHA of the triangle AHC is equal to its interior and remote angle BCA; which (I. 16.) is impossible: wherefore BC is not unequal to EF, that is, it is equal to it. Then, since AB is equal (hyp.) to DE, and the angle B to the angle E: therefore (I. 4.) the base AC is equal to the base DF, the third angle BAC to the third angle EDF, and the triangle ABC to the triangle DEF.* Therefore, if two triangles, &c. If a straight line falling upon two other straight lines, in the same plane, make the alternate angles equal to one another, these two straight lines are parallel. Let the straight line EF, which falls upon the two straight * By the fifth corollary to the 32d proposition of this book, if two angles of one triangle be equal to two angles of another, their remaining angles are also equal. This proposition, therefore, if it be postponed till the student has learned the 32d, may be demonstrated in a single case, the same as either the first or second in Euclid's method, given above: and it may be so postponed witbout impropriety, as the 32d does not depend on it, either directly or indirectly. If the proposition be taken in this order, the enunciation may be as follows:- If two triangles be equiangular to one another, and if a side of the one and a side of the other, which are opposite to equal angles, be equal : then (1.) the remaining sides are equal, each to euch, viz., those which are opposite to equul angles ; and (2.) the triangles are equal. This is the third of the propositions in which triangles are demonstrated to be in every respect equal. They are proved to be such in the fourth proposition, when two sides and the contained angle of the one are respectively equal to two sides and the contained angle of the other : in the eighth, when the three sides of the one are equal to the three sides of the other, each to each ; and in the twentysixth, when two angles and a side of the one are respectively equal to two angles and a side similarly situated of the other. There is only one other case in wbich two triangles are in all respects equal; that is, when a side and the opposite angle of one of them are respectively equal to a side and the opposite angle of the other, and a second side in the one equal to a second side in the other, the angles opposite to these latter sides being of the same kind ; that is, both right angles, both acute, or both obtuse. This proposition, wbich is a case of the 7th of the sixth book, is of little use, except that it completes the theory; but it may be a proper exercise for the student to prove it. It will be seen from these remarks, that two triangles are in every respect equal, when of the sides and angles any three, except the three angles in one of them, are respectively equal to the corresponding sides or angles, in the other, with only the one limitation in the proposition just enunciated, that in it the angles opposite to two equal sides must be of the same kind : and from this it appears, that of the sides and angles of a triangle, three must be given to determine the triangle, and these three cannot be the three angles. Were only the three angles given, the sides might be of any magnitudes whatever. The student may exercise himself in proving this proposition by superposition ; and also by producing BA and BC, instead of cutting off parts of them. H A E B G F D lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD. For, if it be not parallel, AB and CD, being produced, will meet either towards B, D, or towards A, C; let them be produced and meet towards B, D, in the point G. Therefore GEF is a triangle, and its exterior angle AEF is greater (I. 16.) c than the interior and remote angle EFG; but (hyp.) it is also equal to it, which is impossible ; therefore, AB and CD, being produced, do not meet towards B, D. In like manner, it may be demonstrated that they do not meet towards A, C. But those straight lines which meet neither way, though produced ever so far, are parallel (I. def. 11.) to one another. AB therefore is parallel to CD. * Wherefore, if a straight line, &c. PROP. XXVIII. THEOR. E G A B Two straight lines are parallel to one another, (1.) if a straight line falling upon them make the exterior angle equal to the interior and remote, upon the same side of that line; and (2.) if it make the interior angles upon the same side together equal to two right angles. Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and remote angle GHD upon the same side of EF, or make the interior angles on the same side, BGH, GHD, together c equal to two right angles ; AB is parallel to CD. Because the angle EGB is equal (hyp.) to GHD, and EGB equal (I. 15.) to AGH, the angle AGH is equal to GHD; and they are alternate angles; therefore AB is parallel (I. 27.) to CD. Again, because the angles BGH, GHD, are equal (hyp.) to two right angles ; and that AGH, BGH, are also equal (I. 13.) to two right angles ; the angles AGH, BGH, are equal to BGH, D H • In this proof ABG and CDG must be regarded as straight lines. It is not necessary, indeed, to make the actual construction here given ; for the student will see that AB and CD cannot meet on either side, since, if they did, a triangle would thus be formed, and the exterior angle would be (I. 16.) greater than the interior and remote angle, which is contrary to the hypothesis. The alternate angles as they are understood by Euclid, or the interior alternate angles, as they might be called with perhaps more propriety, are the interior remote angles on opposite sides of the line which falls on the other two. AEH and KFD, and also BEH and CFK, may be called exterior alternate angles ; and it is easy to prove that if these be equal, the lines are parallel. GHD; take away the common angle BGH; therefore the remaining angles AGH, GHD are equal ; and they are alternate angles ; therefore AB is parallel (1. 27.) to CD. * Wherefore two straight lines, &c. PROP. XXIX. THEOR. them ; G B If a straight line fall upon two parallel straight lines, (1.) it makes the alternate angles equal to one another ; (2.) the exterior angle equal to the interior and remote upon the same side ; and (3.) the two interior angles upon the same side together equal to two right angles. + Let the straight lines AB, CD be parallel, and let EF fall upon then (1.) the alternate angles AGH, GHD are equal to one another ; (2.) the exterior angle EGB is equal to the interior and remote, upon the same side, GHD; and (3.) the two interior angles ABGH, GHD upon the same side are together c equal to two right angles. For, if AGH be not equal to GHD, one of them must be greater than the other ; let AGH be the greater, and to each of them add angle BGH: therefore the angles AGH, BGH are greater (I. ax. 4.) than the angles BGH, GHD; but the angles AGH, BGH are equal (I. 13.) to two right angles ; therefore the angles BGH, GHD are less than two right angles. But (I. ax. 12.) those straight lines which, with another straight line falling upon them, make the interior angles on the same side less that two right angles, meet together, if continually produced ; therefore the straight lines AB, CD, if produced far enough, will meet : but (I. def. 11.) they never meet, since (hyp.) they are D H н It is easy to prove that if the exterior angles on the same side of the line which falls on the other two, be together equal to two right angles, the straight lines are parallel. This proposition will readily lead us to admit the truth of what is called the 12th axiom. It is here proved, that if the two angles BGH, GHD be together equal to two right angles, A B and CD are parallel, and consequently do not meet. Now, if through G a straight line be drawn in the angle BGH, and consequently making with GH an angle less than BGH, so that the two angles which GH makes with it and with HD are together less than two right angles, we will readily admit that if that line and HD be continually produced they will at length meet : which is the 12th axiom. The truth of this axiom, therefore, will be admitted by any one who has read the Elements thus far, but not, as is the case in respect to the other axioms, by a person who is unacquainted with the principles of geometry. † The first part of this proposition is the converse of the 27th, and the second and third parts are the converses of the first and second parts of the 28th, |