and AC to DF. But the base BC greater than the base EF. The angle BAC is greater than the angle EDF. A B D F For, if the angle BAC be not greater than the angle EDF, it must either be equal to, or less than the angle EDF. The angle BAC is not equal to the angle EDF, because then the base BC would be equal (I. 4) to the base EF: but it is (Hyp.) not equal. Therefore the angle BAC is not equal to the angle EDF. Again, the angle BAC is not less than the angle EDF, because then the base BC would be less (I. 24) than the base EF: but it is (Hyp.) not less. Therefore the angle BAC is not less than the angle EDF. And it was shown that the angle BAC is not equal to the angle EDF. Therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E. D. Corollary. If two triangles have two sides of the one respectively equal to two sides of the other, the base of the one is greater than, equal to, or less than the base of the other, according as the angle opposite to the base of the one is greater than, equal to, or less than the angle opposite to the base of the other. If two triangles have two angles of the one equal to two angles of the other, each to each; and one side equal to one side,-viz., either the sides adjacent to the equal angles, or the sides opposite to equal angles in each ;then their other sides are equal, each to each, and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the two angles ABC, BCA of the one, equal to the two angles DEF, EFD of the other, each to each; viz., ABC to DEF, and BCA to EFD. Also, a side of the one triangle equal to a side of the other. D First, let those sides be equal which are adjacent to the angles that are equal in the two triangles; viz., BC to EF. Then their other sides are equal, each to each; viz., AB to DE, and AC to DF; and the third angle BAC is equal to the third angle EDF. For, if AB be not equal to DE, one of them must be greater than the other. Let AB be the greater of the two. Make BG equal (I. 3) to D F, and join GC. G B C E Because in the two triangles GBC, DEF, BG is equal (Const.) to DE, and BC (Hyp.) to EF, the two sides GB, BC are equal to the two sides DE, EF, each to each. But the angle GBC is equal (Hyp.) to the angle DEF. Therefore the base GC is equal (I. 4) to the base DF, and the triangle GBC to the triangle DEF. And the remaining angles of the one are equal to the remaining angles of the other, each to each; viz., those to which the equal sides are opposite. Therefore the angle GCB is equal to the angle DFE. But the angle DFE is (Hyp.) equal to the angle BCA. Wherefore also the angle BCG is equal (Ax. 1) to the angle BCA, the less to the greater, which is impossible. Therefore the side AB is not unequal to the side DE; that is, AB is equal to DE. And BC is equal (Hyp.) to EF. Therefore the two sides AB, BC are equal to the two sides DE, EF, each to each. And the angle ABC is equal (Hyp.) to the angle DEF. Therefore the base AC is equal (I. 4) to the base DF, and the third angle BAC to the third angle EDF. A D Next, let those sides which are opposite to equal angles in each triangle be equal to one another; viz., AB to DE. Then their other sides are equal; viz., AC to DF, and BC to EF. And the third angle BAC is equal to the third angle EDF. B HC E F For, if BC be not equal to EF, one of them must be greater than the other. Let BC be the greater of the two. Make BH equal (I. 3) to E F, and join A H. Because in the two triangles ABH, DEF, BH is equal (Const.) to EF, and AB to (Hyp.) DE; the two sides A B, BH are equal to the two sides DE, EF, each to each. But the angle A BH is equal (Hyp.) to the angle DEF. Therefore the base A H is equal to the base D F, and the triangle ABH to the triangle DE F. And the remaining angles of the one are equal to the remaining angles of the other, each to each; viz., those to which the equal sides are opposite. Therefore the angle BHA is equal to the angle EFD. But the angle EFD is equal (Hyp.) to the angle BCA. Therefore also the angle BHA is equal (Ax. 1) to the angle B CA; that is, the exterior angle B HA of the triangle A H C is equal to its interior and opposite angle B CA; which is impossible (I. 16). Therefore BC is not unequal to EF; that is, BC is equal to E F. And AB is equal (Hyp.) to D E. Therefore the two sides A B, B C are equal to the two sides DE, EF, each to each. And the angle ABC is equal (Hyp.) to the angle DEF. Therefore the base AC is equal (I. 4) to the base D F, and the third angle BAC to the third angle ED F. Therefore, if two triangles, &c. Q. E. D. The enunciation of this proposition may be thus simplified: If two triangles have two angles of the one, equal to two angles of the other, each to each, and a side of the one equal to a side of the other similarly situated as to the equal angles, the two triangles are equal in every respect. Exercise 1.-In an isosceles triangle, if a straight line be drawn from the angle opposite the base, bisecting the angle, it bisects the base; or, if it bisect the base, it bisects the angle; and in either case, it cuts the base at right angles. Exercise 2.-Through a given point to draw a straight line which shall make equal angles with two straight lines given in position. If a straight line falling on two other straight lines, make the alternate angles equal to each other; these two straight lines are parallel. Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles A E F, EFD, equal to one another. Then A B shall be parallel to CD. For, if AB be not parallel to CD, AB and CD being produced will meet either towards A and C, or towards B and D. Let AB, CD be produced and meet towards B C E B and D, in the point G. Then GEF is a triangle, and its exterior angle AEF is greater (I. 16) than its interior and opposite angle EFG. But the angle AEF is equal (Hyp.) to the angle EFG. Therefore the angle A EF is both greater than, and equal to the angle EFG; which is impossible. Wherefore AB, CD being produced, do not meet towards B, D. In like manner, it may be proved, that they do not meet when produced towards A, C. But those straight lines in the same plane, which do not meet either way, though produced eyer so far, are parallel (Def. 35) to one another. Therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. The angles AEF, EFD, are called alternate angles, or more properly, interior alternate angles, because they are on opposite sides of the straight line EF, and the one has its vertex at E the one extremity of the portion between the parallels, while the other has its vertex at F the other extremity of the same. In the diagram, the crooked lines E BG, FDG, must be considered straight lines, and the figure EFDG B, a triangle, for the sake of the argument. Exercise 1.-If a straight line falling upon two other straight lines, make the exterior alternate angles A GE, FHD (see fig. to next proposition) equal to each other, these two straight lines are parallel. Exercise 2.-If a straight line falling upon two other straight lines, make the two exterior angles on the same side of it, equal to two right angles, these two straight lines are parallel. If a straight line falling upon two other straight lines, make the exterior angle equal to the interior and opposite angle upon the same side of the straight line ; or make the two interior angles upon the same side of it, together equal to two right angles; these two straight lines are parallel to one another. Let the straight line EF, falling upon the two straight lines AB, CD, make the exterior angle EGB equal to the interior and opposite angle GHD upon the same side of EF; or make the two interior angles BGH, GHD on the same side of it, together equal to two right angles. Then AB is parallel to CD. A E Because the angle EGB is equal (Hyp.) to the angle GHD. And the angle EGB is equal (I. 15) to the angle AGH. Therefore the angle AGH is equal (Ax. 1) to the angle GHD; and they are alternate angles. Therefore AB is parallel (I. 27) to CD. G B Again, because the two angles BGH, GHD are together equal (Hyp.) to two right angles. And the two angles AGH, BGH are also together equal (I. 13) to two right angles. Therefore the two angles AGH, BGH are equal (Ax. 1) to the two angles BGH, GHD. Take away from these equals, the common angle BGH. Therefore the remaining angle AGH is equal (Ax. 3) to the remaining angle GHD; and they are alternate angles. Therefore AB is parallel (I. 27) to CD. Wherefore, if a straight line, &c. Q. E. D. The twelfth axiom will now be admitted as a corollary to this proposition; espe cially when Prop. XVII. and the note added to that axiom are taken into account. PROP. XXIX. THEOREM. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another; the exterior angle equal to the interior and opposite angle upon the same side of the straight line; and the two interior angles upon the same side of it together equal to two right angles. E Let the straight line EF fall upon the two parallel straight lines A B, CD. The two alternate angles AGH, GHD are equal to one another. The exterior angle EGB is equal to the interior and opposite angle GHD upon the same side of the straight line EF. And the two interior angles BGH, GHD upon the same side of it are together equal to two right angles. A B D H F For, if the alternate angles AGH, GHD be not equal, one of them must be greater than the other. Let the angle AGH be greater than the angle GHD. To each of these unequals, add the angle BGH. Then the two angles AGH, BGH, are greater (Ax. 4) than the two angles BGH, GHD. But the two angles AGH, BGH, are equal (I. 13) to two right angles. Therefore the two angles BGH, GHD, are less than two right angles. But those straight lines, which with another straight line falling upon them, make the two interior angles on the same side less than two right angles, will meet together (Ax. 12) if continually produced. Therefore the two straight lines A B, CD, if produced far enough, will meet. But they never meet, since (Hyp.) they are parallel. Therefore the angle AGH is not unequal to the angle GHD; that is, the angle AGH is equal to the angle GHD. Again, the angle AGH is equal (I. 15) to the angle EGB. Therefore the angle EGB is equal (Ax. 1) to the angle GHD. Lastly, to each of these equals, add the angle BGH. Then the two angles EGB, BGH, are equal (Ax. 2) to the two angles BGH, GHD. But the two angles EGB, BGH are equal (I. 13) to two right angles. Therefore also the two angles BGH, GHD are equal (Ax. 1) to two right angles. Wherefore, if a straight line, &c. Q. E. D. Corollary 1.-If two angles have their legs parallel each to each, and proceeding from their vertices in the same directions, they are equal to each other. Corollary 2.-If two angles be equal to each other, and a leg of the one be parallel to a leg of the other, their remaining legs are parallel. Exercise.-If a straight line be drawn perpendicular to one of two parallel straight, lines, it is also perpendicular to the other. PROP. XXX. THEOREM. Straight lines which are parallel to the same straight line are parallel to each other. Let the straight lines AB, CD, be each of them parallel to EF. Then AB is also parallel to CD. Draw the straight line GHK cutting the three straight lines A B, EF, and CD. Because the straight line GHK cuts the parallel straight lines AB, EF, the angle AGH is equal (I. 29) to the alternate angie G HF. Again, because the straight line GHK cuts the parallel straight lines EF, CD, the exterior angle GHF is equal (I. 29) to the interior angle H K D. But it was proved that the angle AG H is equal to the angle GHF. Therefore the angle AGH is equal (r. 1) to the angle G K D; and these are Ealternate angles. Therefore AB is parallel (1. 27) to CD. Wherefore, straight lines which are de Q. E. D. The student should prove the case of this proposition, A H K when the straight line KF is not between the straight lines A B, CD, but on either side of both. PROP. XXXI. PROBLEM. Daher sirvagi a given point a straight line parallel to a given raight line. Let A be the given point, and B C the given straight line. It is required to draw, through the point A, a straight line parallel to the straight line BC E In the straight line BC take any point D, and form A IX At the point in the straight line AD, Pate & 20 h e PAP equal to the angle B APC Polar the straight dr BARF. Tâca A. A F es the two straight lines EF, BC, FAN ADC equal to one another. FC Wherefore, through the given bez dara parallel to the given q??! @\\\\, \ 'de neu Paste des hul i zessary, if the 11th r the assetetion and demonstration, ppp peut de 12 say be used instead of the 27th Ne, ne bo dog de ase read ngà, and having their bases He op jy die vry poly the le is bisected in that |