its exterior angle AEF is greater (16. 1.) than the interior and opposite angle EFG ; but it is also equal to it, which is impossible; therefore, AB А E B and CD being produced, do not meet towards B, G D. In like manner it may be demonstrated that they с F D do not meet towards A, C; but those straight lines which meet neither way, though produced ever so far, are parallel (30. Def.) to one another. AB therefore is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. PROP. XXVIII. THEOR. If a straight line falling upon two other straight lines makes the exterior angle equal to the interior and opposite upon the same side of the line; or makes the interior angles upon the same side together equal to two right angles; the two straight lines are parallel to one another. Let the straight line EF, which falls upon the two straight lines AB, CD, make the exterior angle EGB equal to GHD, the interior and opposite angle upon the E same side ; or let it make the interior angles on the same side A G B BGH, GHD together equal to two right angles ; AB is parallel to CD. Because the angle EGB is equal to the angle GHD, and с Н. D also (15. 1.) to the angle AGH, the angle AGH is equal to the F angle GHD; and they are the alternate angles ; therefore AB is parallel .(27. 1.) to CD. Again, because the angles BGH, GHD are equal (By Hyp.) to two right angles, and AGH, BGH, are also equal (13. 1.) to two right angles, the angles AGH, BGH are equal to the angles BGH, GHD : Take away the common angle BGH ; therefore the remaining angle AGH is equal to the remaining angle GHD; and they are alternate angles ; therefore AB is parallel to CD. Wherefore, if a straight line, &c. Q. E. D. PROP. XXIX. THEOR. If a straight line fall upon two parallel straight lines, it makes the alter nate angles equal to one another; and the exterior angle equal to the interior and opposite upon the same side ; and likewise the two inierior 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 are equal to one another ; and d the exterior angle EGB is equal to the interior and opposite, upon the same side, GHD; and the two interior angles BGH, GHD upon the same side are together equal to two right angles. For if AGH be not equal to GHD, let KG be drawn making the angle KGH equal to GHD, and produce KG to L; then KL will be ; parallel to CD (27. 1.) ; but E AB is also parallel to CD; L. therefore two straight lines Α. G are drawn through the same -В point G, parallel to CD, and K yet not coinciding with one another, which is impossible C D (11. Ax.). The angles AGH, H GHD therefore are not unequal, that is, they are equal F to one another. Now, the angle 'EGB is equal to AGH (15. 1.); and AGH is proved to be equal to GHD; therefore EGB is likewise equal to GHD; add to each of these the angle BGH; therefore the 'angles EGB, BGH are equal to the angles BGH, GHD; but EGB, BGH are equal (13. 1.) to two right angles; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Q. E. D. Cor. If two lines KL and CD make, with EF, the two angles KGH, GHC together less than two right angles, KG and CH will meet on the side of EF on which the two angles are that are less than two right angles. For, if not, KL and CD are either parallel, or they meet on the other side of EF ; but they are not parallel ; for the angles KGH, GHC would then be equal to two right angles. Neither do they meeť on the other side of EF ; for the angles LGH, GHD would then be two angles of a triangle, and less than two right angles ; but this is impossible ; for the four angles KGH, HGL, CHG, GHD are together equal to four right angles (13. 1.), of which the two, KGH, CHG are by supposition less than two right angles; therefore the other two, HGL, GHD are greater than two right angles. Therefore since KL and CD are not parallel, and since they do not meet towards L and D, they must meet if produced towards K and C. PROP. XXX. THEOR. Straight lines which are parallel to the same straight line are parallel to one another. Let AB, CD, be each of them parallel to EF; AB is also parallel to CD. Let the straight line GHK cut AB, EF, CD; and because GHK cuts the parallel straight lines AD, EF, the angle AGH is equal PROP. XXXÍ. PROB. 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 res quired to draw a straight line through the point A, parallel to A the straight line BC. T T In BC take any point D, and join AD; and at the point A, i in the straight line AD, make B D 19 d (23, 1.) the angle DAE equal to the angle ADC ; and produce the straight line EA tor. Because the straight line AD, which meets the two straight lines BC, EF, makes the alternate angles EAD, ADC equal to one another, EF is parallel (27. 1.) to BC. Therefore the straight line EAF is drawn through the given point A parallel to the given straight line BC Which was to be done. PROP. XXXII. THEOR. 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 is equal to the two interior and opposite angles CAB, ABC ; and the three interior angles of the triangle, viz. C' A E AB is parallel to CE and AC meets them, the alternate angles BAC, ACE are equal (29. 1.). Again, because AB is B ܪ parallel to CE, and BD falls upon them, the exterior angle ECD ig equal to the interior and opposite angle ABC ; but the angle ACE was shewn to be equal to the angle BAC ; therefore the whole exteriof angle ACD is equal to the two interior and opposite angles CAB, ABC; to these angles add the angle ACB, and the angles ACD, ACB are equal to the three angles CBĂ, BAC, ACB ; but the angles ACD, ACB are equal (13. 1.) to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D. Cor. 1. All the interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides, wanting four right angles. For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure ; and the same angles are equal to the angles of 1 the figure, together with the angles at the point F, which is the common vertex of the triangles : that is, (2. Cor. 15. 1.) together with four right angles. Tberefore, twice as many right angles as the figure has sides, are equal to all the angles of the figure, together with four right angles, that is, the angles of the figure are equal to twice as many right angles as the figure has sides, wanting four. Cor. 2. All the exterior angles of any rectilineal figure are tegether equal to four right angles. Because every interior angle ABC, with its adjacent exterior ABD, is equal (13. 1.) to two right angles ; therefore all the interior, together with all А the exterior angles of the figure, are equal to twice as many right angles as there are sides of the fi C gure; that is, by the fore- D B going corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal to four right angles. 21 PROP. XXXIII. THEOR. 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, CD, be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD; AC, BD are also equal and parallel. Join BC; and because AB is pa- A B rallel to CD, and BC meets them, the alternate angles ABC, BCD are equal (29. 1.) ; and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC are equal to the two С D DC, CB ; and the angle ABC is equal to the angle BCD; therefore the base AC is equal (4. 1.) to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles (4. 1.), each to each, to which the equal sides are opposite ; 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, AC is parallel (27. 1.) to BD; and it was shewn to be equal to it. Therefore, straight lines, &c. Q. E. D.D PROP. XXXIV. THEOR. The opposite sides and angles of a parallelogram are equal to one ano ther, and the diameter bisects it, that is, divides it into two equal parts. a N. B. A Parallelogram is a four-sided figure, of which the opposite sides are parallel ; and the diameter is the straight line joining two of its opposite angles. Let ACDB be a parallelogram, of which BC is a diameter ; the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it. Because AB is parallel to CD, and А. B BC meets them, the alternate angles ABC, BCD are equal (29. 1.) to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal (29. 1.) to one another ; wherefore C D the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and the side BC, which is adjacent to these equal angles, common to the iwo triangles ; therefore their other sides are equal, each to each, and 42 |