Hence CD meet, .. ZAGH not unequal to GHD. and <EGB= <GHD, also ZAGH or < AGK = Z GHF, 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 = 4 GKD. .. 4 AGK = 4 GKD. PROOF.-Because GHK cuts the par-B allel 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 any point D, and join AD. 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. A F Make < DAE= < 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). Make Then 4 BAC= ZACE, 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). 4 ABC. .. 4 ACD 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. E A D And, by the preceding proposition, the angles of each triangle are equal to two right angles. Therefore all the angles of the triangles Bare equal to twice as many right angles is 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, arc 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 A B 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 4 BCD. 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; C B Therefore the base AC is equal to the base BD (I. 4), .. AC == BD, and < ACB = 4 CBD. < ABC = < BCD, and 4 CBD. Therefore the angle ACB is equal to the angle CBD. And because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another; Therefore AC is parallel to BD (I. 27); and it was shown to be equal to it. Therefore, the straight lines, &c. Q. E. D. Proposition 34.-Theorem. The opposite sides and angles of a parallelogram are equal to one another, and the diagonal bisects the parallelogram—that is, divides it into two equal parts. Let ACDB be a parallelogram, of which BC is a diagonal; The opposite sides and angles of the figure shall be equal to one another, And the diagonal BC shall bisect it. PROOF.-Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal to one another (I. 29); B D Because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal to one another (I. 29); Therefore the two triangles ABC, BCD have two angles, ABC, BCA in the one, equal to two angles, BCD, CBD in the other, each to each; and the side BC, adjacent to the equal angles in each, is common to both triangles. Therefore the other sides are equal, each to each, and the third angle of the one to the third angle of the other— namely, AB equal to CD, AC to BD, and the angle BAC to BD, BAC the angle CDB (I. 26). AB = CD, AC = = 4 CDB, and And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, Therefore the whole angle ABD is equal to the whole ACD, angle ACD (Ax. 2). .. 4 ABD And the angle BAC has been shown to be equal to the angle BDC; therefore the opposite sides and angles of a parallelogram are equal to one another. Also the diagonal bisects it, |