• Intenc, 1. and ACD are" together equal to two right angles. And because #. CAB and ACD aret together less than two right angles, they are: tinterc. 1. less than CAF and ACD. Wherefore, by taking away the *:::. a common angle ACD, the angle CAB is less than CAF. Cor. 8. And because CG ist equal to CA, the angle CAG ist equal lot. to CGA.; and because these angles CAG, CGA are also" together *i.js E. not less than the exterior angle ACE, CAG is not less than half #. 1. the angle ACE, or than half the angle CAF which ist equal to it. Cor. 12. In GD takef GH equal to GA, and join HA; and it may be shown #I. 3. in like manner that the angle GAH is not less than half the angle AGE. But the angle AGE is equal to CAG, which is not less than half CAF and consequently not less than the remaining angle GAF; wherefore the angle AGE is not less than GAF, and GAH (which is not less than half AGE) is not less than half GAF. And in like manner take HI equal to HA, and join IA; and so on. Wherefore, because from the angle CAF is cut off not less than its half, and from the remainder not *INTEnc. l. less than its half, and so on; there will” at length remain an or.sangle, as KAF, less than FAB, and AB being prolonged willf cut ... the perimeter of the triangle CAK, in some point between C and K; that is to say, it will meet CD. And by parity of reasoning, the like may be proved in every other instance. Wherefore, universally, if two straight lines which are in the same plane are met by a third, and the two interior angles &c. Which was to be demonstrated. PROPOSITION XXIX. THEOREM.–If a straight line falls upon two parallel straight lines, it makes the alternate angles equal to one another. Let the straight line EF fall upon the E. parallel straight lines AB, CD; the alter- G B nate angles GHC, HGB are equal to one G H D another. F For, if the angle GHC be not equal to HGB, one of them must be the less. Let HGB be assumed to be the less. Wherefore, since HGB is less than GHC, if each be added to the tintenc, 1. angle GHD, the angles HGB and GHD must bef together less 9; than the angles GHC and GHD. But GHC and GHD are" *I. 28 F. together equal to two right angles; therefore HGB and GHD upon two parallel straight lines, &c. Which was to be demonstrated. CoR. I. If a straight line falls upon two parallel straight lines, it makes the exterior angle equal to the interior and opposite on the same side of the line; and makes also the two interior angles on the same side, together equal to two right angles. For First ; because (by the Prop. above) it makes the alternate angles GHC, HGB equal to one another, and HGB ist equal to EGA, the angle EGA ist equal to GHC. And in the same way, because the alternate angles GHD, HGA are equal, might be shown that EGB and GHD are equal. Secondly ; because the angle GHC is equal to HGB, add to each the angle GHD, and the angles GHC and GHD will be" together equal to HGB and GHD. But GHC and GHD aret together equal to two right angles; therefore HGB and GHD aret together equal to two right angles. And in the same way, by adding to GHD and HGA the angle GHC, might be shown that HGA and GHC are together equal to two right angles. CoR. 2. Through a point in one of two parallel straight lines, no other straight line can be drawn in the same plane, which (all the straight lines being continually prolonged both ways) shall not meet the other of the two parallel straight lines. For if through any point in CD, as H, any straight line other than CD be drawn, it will make with a line drawn from H to any point G in AB, the two interior angles on one side or on the other together less than two right angles ; therefore (on the lines being prolonged) it will” meet AB. f I.29. *I.29.Cor.1. #INTERC. l. Cor. 5. # INTERC. l. Cor. 1. PROPOSITION XXX. THEOREM.–Straight lines nhich are parallel to the same straight line (the nhole being in the same plane), are parallel to one another. Let AB, CD be each of them parallel to EF; all the three being in the same plane. AB shall be parallel to CD. A Let thestraight line GKH cut AB, CD, EF. c Because GKH cuts the parallel straight E lines AB and EF, the alternate angles AGH, GHF are” equal to one another. Again, because thestraight line GKH cuts the parallel straight lines CD and EF, the exterior angle GKD ist equal to the interior and opposite on the same side, GHF. And it was shown that the angle AGH is equal to GHF; therefore the angle AGH is: equal to GKD; and they are alternate angles; therefore AB is" parallel to CD. And by parity of reasoning, the like may be proved in every other instance. Wherefore, universally, straight lines which are parallel to the same straight line &c. Which was to be demonstrated. Scholium.—The same is true when the straight lines are not all in the same plane. But this is not demonstrated by Euclid till Book XI. PROPOSITION XXXI. THEoREM.—If a side of any triangle be prolonged, the exterior angle is equal to the sum of the two interior and opposite angles. Let ABC be a triangle, and let one of its sides, as BC, be produced to D. The ex- A E terior angle ACD shall be equal to the two interior and opposite angles CAB, ABC. Because CE is parallel to BA, and CA falls upon them, the alternate angles ACE, CAB are: equal. Also, because DB falls upon them, the exterior angle ECD is" equal to ABC the interior and opposite on the same side of the line. Therefore the sum of ACE and ECD ist equal to the sum of CAB and ABC. But the sum of ACE and ECD is the angle ACD; therefore ACD is: equal to the sum of CAB and ABC. And by parity of reasoning, the like may be proved in every other instance. Wherefore, universally, if a side of a triangle be prolonged, &c. Which was to be demonstrated. PROPOSITION XXXII. THEOREM.–The three interior angles of every triangle are equal to two right angles. Let ABC be a triangle. The three interior A. angles CAB, ABC, and BCA, are together equal to two right angles. . *INTERc.12. Prolong” one side of the triangle; as BC. ##. Because the exterior angle ACD ist equal to B C D the sum of the two interior and opposite angles ABC and CAB ; HNTERc. 1. add to each the angle BCA, and the sum of ACD and BCA is: Cor. 4. equal to the sum of the three angles ABC, CAB, and BCA. * I. 12. But the sum of ACD and BCA is" equal to two right angles, go. 1. Therefore the sum of the three angles ABC, CAB, and BCA ist “ equal to two right angles. And by parity of reasoning, the like may be proved in every other instance. Wherefore, universally, the three interior angles of a triangle are equal to two right angles. Which was to be de monstrated. CoR. 1. All the interior angles of any plane rectilinear figure that incloses a space, are equal to twice as many right angles as the figure has sides, diminished by four right angles. For First ; if the figure is one, as ABCDE, TX that can be divided into as many triangles as the C figurehas sides, by drawing straight lines from one point F within the figure to each of its angular points. All the angles of these triangles (by the A. B preceding Proposition) are together equal to twice as many right angles as there are triangles; that is, to twice as many right angles as the figure has sides. But the angles at the point F are t1.12.Cor.2, together; equal to four right angles. Therefore the other angles of the triangles, (which are together equal to all the interior *INTEnc. 1. angles of the figure), are together" equal to twice as many right Cor. 7. angles as the figure has sides, diminished by four right angles. *INTERC. l. See Note. +I. 12. : INTERc. 1. Cor. 4. *INTERC. I. +INTERC. 1. Cor. 7. Secondly; if the figure be such as cannot be divided into triangles by lines drawn from one point within the figure, it can be divided into smaller rectilinear figures each severally capable of being so divided. And because the interior angles of each of such smaller figures are equal to twice as many right angles as the figure has sides, diminished by four right angles, (or, which is the same thing, to twice as many right angles as the sides of the figure with the subtraction of two sides); the interior angles of all these smaller figures are together equal to twice as many right angles as is the number of the sides of all those figures after subtraction of two sides from each. And this number after subtraction, is equal to the number of sides of the whole or composite figure with the subtraction of two; for every straight line that makes division between the smaller figures, constituted and was counted as a side of two of those figures, and the number of the divisions is less by one than the number of the figures. Wherefore the interior angles of all the smaller figures, are" together equal to twice as many right angles as is the number of sides of the whole or composite figure with the subtraction of two; or, which is the same thing, to twice as many right angles as the whole or composite figure has sides, diminished by four right angles. CoR. 2. In any plane rectilinear figure A For because every interior angle ABC, with its adjacent exterior ABD, isł equal to two right angles; 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 (by Cor. 1.) all the interior angles are together equal to twice as many right angles as the figure has sides, diminished by four right angles; and (by adding four right angles to each), all the interior angles together with four right angles, are: equal to twice as many right angles as the figure has sides. Therefore all the interior together with all the exterior angles of the figure, are" equal to all the interior angles of the figure together with four right angles; and by taking away the interior angles from both, all the exterior angles of the figure aret equal to four right angles. |