PROP. XXV. THEOR. If two triangles (BAC, DEF) have two sides (AB, AC) of the one respectively equal to two sides (DE, DF) of the other, but their bases unequal, the angle subtended by the greater base (BC) of the one, must be greater than the angle subtended by the less base (EF) of the other. The angle BAC subtended by the greater base must be equal to, or less, or greater than the angle EDF subtended by the less base; BAC is not equal to but EDF, for if it were, the base BC would be equal to EF E D If two triangles (ABC, DEF) have two angles (B, C) of the one respectively equal to two angles (E, F) of the other, and also a side in the one equal to a side similarly situated in the other, that is to say, either lying between the equal angles or subtending equal angles; then must the remaining sides and angle of the one triangle be respectively equal to the remaining sides and angle of the other. First, let the given equal sides be BC and EF, which lie between the equal angles; then the sides subtending the equal angles in the two triangles shall be respectively equal. For if two of those sides, as BA and ED subtending equal angles be not equal, then one of them must be greater than the other; and from ED, which is assumed to be the greater, cut off a part EG equal to BA, and conterminous with EF, and join GF. Then, since the triangles BAC, EGF have BA=EG, BC= EF, and ABC = / GEF (Hyp.), they are equal (Prop. 4), and have ACB = / GFE ; but / ACB = ZDFE (Hyp.); therefore / GFE = DFE, a part equal to the whole, which is absurd (Ax. 9). Therefore of the sides BA, ED, neither is greater than the other, but they are equal: and since the triangles ABC, DEF have BA = ED, BC = EF, and < ABC = /DEF, they have their remaining sides and angles equal (Prop. 4), viz. AC=DF and / BAC=/ EDF. B E G F Secondly, let the given equal sides be sides subtending equal angles, as AC and DF; now in this case, the sides CB, FE lying between the equal angles shall be also equal. For suppose it possible that one of them, as FE, be greater than the other, then cut off from FE a part FG equal to CB and conterminous with DF, and join DG. Then, since AC = DF, CB = FG, and ▲ ACB = / DFE (Hyp.), the triangles ACB and DFG are equal (Prop. 4), and ABC=/DGF; but /ABC=/DEF (Hyp.). Therefore DGF=/ DEF; but / DGF is exterior to, and therefore also greater than ▲ DEF (Prop. 16), which is absurd. Therefore the sides CB, FE are not unequal; and since they are equal, and also AC=DF, and ≤ ACB=/ DFE, the triangles BAC and EDF have their bases and remaining angles equal, viz. AB=DE, and ▲ BAC=≤ EDF (Prop. 4). PROP. XXVII. THEOR. If a straight line (EF) meeting two other straight lines, (AB and CD) makes with them the alternate angles (AEF and EFD, BEF and EFC) equal, these two straight lines are parallel. A F E/B G For if the lines AB, CD be not parallel, they shall meet when produced. Suppose that they meet in the point G, then EGF is a triangle, and its side GE being produced, the exterior angle AEF is greater than the internal remote angle EFG (Prop. 16); but also AEF=/ EFD (Hyp.) which is absurd. Therefore the straight lines AB, CD cannot meet in G; and in like manner it may be demonstrated that they cannot meet in any other point: they are therefore parallel to each other (Def. 10). PROP. XXVIII. THEOR. If a straight line (EF) cutting two other straight lines (AB and CD), makes the external equal to the internal and opposite angle, at the same side of the cutting line (viz. EGB = / GHD, or ▲ EGA = ▲ GHC), or if it makes the two internal angles at the same side (viz. BGH and GHD, or AGH and GHC), together equal to two right angles, those two straight lines are parallel. A C E G B First, if the external equals the internal and opposite angle, or EGB = / GHD; then because ▲ EGB=/ AGH (Prop. 15), the alternate angles AGH, GHD are also equal; and therefore the straight lines AB, CD are parallel (Prop. 27). Secondly, if BGH + ≤ GHD = two right angles, then, since likewise BGH+▲ AGH = two right angles (Prop. 13), if ▲ BGH be taken from both, there will remain / GHD=AGH (Ax. 3); but these are alternate angles, and therefore the straight lines AB, CD are parallel (Prop. 27). H D F A straight line (EF) falling on two parallel straight lines (AB and CD), makes the alternate angles equal to one another; and also the external equal to the internal and opposite angle on the same side; and the two internal angles on the same side together equal to two right angles. K H E Ꮮ B For if the alternate angles AGH, GHD be not equal, draw the line GK, making the angle HGK equal to GHD (Prop. 23), and produce KG to L; then KL must be parallel to CDA(Prop. 27); and therefore the two straight lines AB and KL intersecting in the point c G, are parallel to the same straight line, which is impossible (Ax. 11). Therefore the alternate angles AGH, GHD are not unequal, and since they are equal, and LAGH=/EGB (Prop. 15), the external angle EGB equals the internal opposite angle GHD on the same side; if to both F be added BGH, then EGB+/BGH=/BGH+/GHD; but the former equal two right angles (Prop. 13), therefore the latter also, that is to say, the two internal angles at the same side, are equal to two right angles. PROP. XXX. THEOR. Straight lines (AB, CD) which are parallel to the same straight line (EF), are parallel to one another. Let the straight line GK cut the lines AB, EF and CD ; then, since AB and EF are parallel, ▲ AGH = ZGHF (Prop. 29); and in like manner, since A EF and CD are parallel, GHF = HKD; E H therefore AGH=/HKD; but these are the CK alternate angles made by the cutting line GK G/ B with the straight lines AB and CD, which are therefore parallel (Prop. 27). From a given point (c) to draw a straight line parallel to a given straight line (AB). D с E In the given straight line AB take any point F, join FC, and at the point C in the line FC make the angle FCD equal to /CFB (Prop. 23), duce DC to E; the straight line DE shall be parallel to AB. For DCF=/ CFB (Const.), A and pro F B but they are alternate angles; therefore the straight lines DE, AB are parallel (Prop. 27). If a side of any triangle (ABC) is produced, the external angle is equal to the two internal remote angles taken together: and the three internal angles of every triangle, are together equal to two right angles. If the side BA be produced towards F, the external angle FAC shall be equal to the two internal angles B and C. For, F G from the point A, draw AG parallel to BC (Prop. 31); and then ZGAC=/C, because they are alternate angles; and / FAG=/ B, because the former is the external, and the latter the internal opposite A angle (Prop. 29); therefore the external angle FAC =2B+C; let BAC be added to both; then / FAC+/ BAC=/ BAC+≤B+≤C; but Ls FAC, BAC=two right angles (Prop. 13); therefores BAC, B, and C, which are the internal angles of the triangle, are also equal to two right angles. B C COR. 1.-If two triangles have two angles in the one, together equal to two angles in the other, they have their remaining angles also equal. COR. 2.-If one angle of a triangle be a right angle, the other two are together equal to a right angle. COR. 3.-In an equilateral triangle, each angle is a third of two right angles, and is therefore equal to two-thirds of a right angle. COR. 4.-All the internal angles of any rectilinear figure are equal to twice as many right angles (deducting four) as the figure has sides. B с E For, if from any point F, taken within the figure, the straight lines FA, FB, FC, &c. be drawn to the extremities of the sides, the figure will be thus divided into as many triangles as the figure has sides, and the angles of each triangle are equal to two right angles. All the angles, therefore, of the triangles into which the A figure is divided, are equal to twice as many right angles as the figure has sides. But of these, the angles round the point F are equal to four right angles (Prop. 13, Cor.): if these be taken away, therefore, the remaining angles, which are those of the figure, will be equal to twice as many right angles (deducting four) as the figure has sides. COR. 5.-All the external angles of any rectilinear figure, are together equal to four right angles. For, every external angle ABC, with its internal adjacent angle DBC, is equal to two right angles. Therefore, all the external, A with all the internal angles of the figure, are together equal to twice as many right angles B D as the figure has sides; that is to say, according to the |