Therefore the angles BGH, AGH are equal to the angles 4 BGH BGH, GHD (Ax. 1). Take away the common angle BGH. + AGH = BGH +4 GHD. = 4 GHD Therefore the remaining angle AGH is equal to the remain- .. < AGH ing angle GHD (Ax. 3), and they are alternate angles. Therefore AB is parallel to CD (I. 27). Therefore, if a straight line, &c. Q. E. D. Proposition 29.-Theorem. If a straight line fall upon two parallel straight lines, it makes the alternate angles equal to one another, and the exterior angle equal to the interior and opposite upon the same side; and also the two interior angles upon the same side together equal to two right angles. Let the straight line EF fall upon the parallel straight lines AB, CD; The alternate angles AGH, GHD shall be equal to one another. The exterior angle EGB shall be equal to GHD, the interior and opposite angle upon the same side; And the two interior angles on the same side BGH, GHD shall be together equal to two right angles. For if AGH be not equal to GHD, one of them must be greater A E G than the other. Let AGH be the greater. PROOF.-Then the angle AGH is greater than the angle GHD; to each of them add the angle BGH. Therefore the angles BGH, AGH are greater than the angles BGH, GHD (Ax. 4). But the angles BGH, AGH are together equal to two right angles (1. 3). <AGH > GHD. (supposc.) ..< BGH Therefore the angles BGH, GHD are less than two right + GHD angles. But if a straight line meet two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles, these straight lines being <two right angles. Hence CD meet, .. ZAGH, not unequal to GHD. and also 4 BGH + / GHD = two right angles. ZAGH or continually produced, shall at length meet on that side on which are the angles which are less than two right angles (Ax. 12); Therefore the straight lines AB, CD will meet if produced far enough. But they cannot meet, because they are parallel straight lines (Hyp.); Therefore the angle AGH is not unequal to the angle GHD-that is, it is equal to it. But the angle AGH is equal to the angle EGB (I. 15); Therefore the angle EGB is equal to the angle GHD (Ax. 1). Add to each of these the angle BGH. Therefore the angles EGB, BGH, are equal to the angles BGH, GHD (Ax. 2). But the angles EGB, BGH, are equal to two right angles (I. 13). Therefore also BGH, GHD, are equal to two right angles (Ax. 1). Therefore, if a straight line, &c. Q. E. D. Proposition 30.-Theorem. Straight lines which are parallel to the same straight lines are parallel to one another. Let AB, CD be each of them parallel to EF; CONSTRUCTION.-Let the straight line GHK cut AB, A and H E <GHF = < GKD. ..Z AGK =/ GKD. PROOF.-Because GHK cuts the parBallel straight lines AB, EF, the angle AGH is equal to the angle GHF (I. 29). Again, because GK cuts the parallel straight lines EF, CD, the angle GHF is equal to the angle GKD (I. 29). And it was shown that the angle AGK is equal to the angle GHF; Therefore the angle AGK is equal to the angle GKD (Ax. 1), and they are alternate angles; Therefore AB is parallel to CD (I. 27). Proposition 31.-Problem. To draw a straight line through a given point, parallel to a given straight line. Let A be the given point, and BC the given straight line. It is required to draw a straight line through the point A, parallel to BC. CONSTRUCTION.-In BC take point D, and join AD. any E At the point A, in the straight line AD, make the angle DAE equal to the B angle ADC (I. 23). Produce the straight line EA to F. D A F Make DAE= Z ADC. alternate PROOF.-Because the straight line AD, which meets the They are two straight lines BC, EF, makes the alternate angles EAD, angles. ADC equal to one another; Therefore EF is parallel to BC (I. 27). Therefore, the straight line EAF is drawn through the given point A, parallel to the given straight line BC. Q. E. F. Proposition 32.-Theorem. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to D; The exterior angle ACD shall be equal to the two interior and opposite angles CAB, ABC; And the three interior angles of the triangle-namely, ABC, BCA, CAB, shall be equal to two right angles. CONSTRUCTION.-Through the B point C, draw CE parallel to AB (I. 31). E Make Then < BAC= < ACE, and PROOF.-Because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (I. 29). Again, because AB is parallel to CE, and BD falls upon < ECD = them, the exterior angle ECD is equal to the interior and opposite angle ABC (I. 29). < ABC. .. < ACD = Add BAC But the angle ACE was shown to be equal to the angle BAC; Therefore the whole exterior angle ACD is equal to the ABC. two interior and opposite angles BAC, ABC (Ax. 2). To each of these equals add the angle ACB. ACB. Therefore the angles ACD, ACB are equal to the three angles CBA, BAC, ACB (Ax. 2). But the angles ACD, ACB are equal to two right angles (T. 13); Therefore also the angles CBA, BAC, ACB are equal to two right angles (Ax. 1). Therefore, if a side of any triangle, &c. Q. E. D. COROLLARY 1.-All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. For any rectilineal figure ABCDE can, by drawing straight lines from a point F within the figure to each angle, be divided into as many triangles as the figure has sides. D B And, by the preceding proposition, the angles of each triangle are equal to two right angles. Therefore all the angles of the triangles are equal to twice as many right angles as there are triangles; that is, as there are sides of the figure. But the same angles are equal to the angles of the figure, together with the angles at the point F; And the angles at the point F, which is the common vertex of all the triangles, are equal to four right angles (I. 15, Cor. 2); Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. COROLLARY 2.-All the exterior angles of any rectilineal figure are together equal to four right angles. The interior angle ABC, with its adjacent exterior angle ABD, is equal to two right angles (I. 13); Therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides. But all the interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides (I. 32, Cor. 1); Therefore all the interior angles, together with all the exterior angles, are equal to all the interior angles and four right angles (Ax. 1). D Take away the interior angles which are common ; Therefore all the exterior angles are equal to four right angles (Ax. 3). Proposition 33.-Theorem. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are also themselves equal and parallel. Let AB and CD be equal and parallel straight lines joined towards the same parts by the straight lines AC and BD ; AC and BD shall be equal and parallel. CONSTRUCTION. Join BC. PROOF.-Because AB is parallel to CD, and BC meets ABC = them, the alternate angles ABC, BCD are equal (I. 29). Because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides A AB, BC are equal to the two sides DC, CB, each to each; And the angle ABC was proved to be equal to the angle BCD; B Therefore the base AC is equal to the base BD (I. 4), < BCD. . AC = BD, |