But it was proved that the BFL is equal to the FLC. Therefore the EFA is also equal to the In a similar way it might be shown that the equal to the LF B, and the HLC to the FLC. HLD is LFA. PROPOSITION XXX. Straight lines which are parallel to the same straight line are parallel to each other. To prove this proposition we must know, 1. That magnitudes which are equal to the same are equal to one another. (Ax. I.) 2. That if two straight lines are parallel, and are cut by a third straight line, the alternate s so formed are equal. (Prop. XXIX.) · 3. That if two straight lines are parallel, and are cut by a third, the exterior is equal to the interior and opposite on the same side of the intersecting line. (Prop. XXIX.) 4. That if two straight lines are crossed by a third, and the exteriorso formed is equal to the interior and opposite on the same side of the intersecting line, those straight lines are parallel. (Prop. XXVIII.) Case 1. Suppose the two straight lines A B and C D are each parallel to EF. We have to prove that A B and CD are parallel to each other. G H A B EF are parallel and the line G M cuts them, the exterior GHA is equal to the interior and opposite ZHLE. Again, because CD and EF are parallel, the exterior HKC is equal to the interior and opposite KLE. Therefore, since the s GHA and HKC are both equal to the KLE, they are equal to each other. Now the 8 GHA and HKC are the exterior and interior opposites, formed by the intersection of the two lines A B and CD by the line G M. But if two straight lines are crossed by a third, and the exteriorso formed is equal to the interior and opposite on the same side of the intersecting line, those straight lines are parallel. (Prop. XXVIII.) Therefore the lines A B and CD are parallel. the line G M intersects them, the exterior GHA is equal to the interior and opposite HLE. Again, because EF and CD are parallel, and the line GM intersects them, the exteriorHLE is equal to the interior and opposite LKC. Therefore, since the s G H A and L K C are both equal to the HLE, they are equal to each other. We have, therefore, two straight lines, A B and CD, cut by a third, so that the exterior thus formed (namely G HA) is equal to the interior and opposite on the same side of the intersecting line (namely, the HKC). It follows, therefore, according to the 28th proposition that these two right lines A B and C D are parallel to each other. PROPOSITION XXXI, To draw a straight line through a given point parallel to a given straight line. For the construction requisite in this proposition we must be able, 1. To join two given points by a straight line. 2. From a given point in a given straight line to That if two straight lines are met by a third, and the alternates so formed are equal to one another, those two straight lines are parallel. (Prop. XXVII.) Let AB be the given straight line, and D the given point. We have to draw through the point D a line parallel to the given line A B. In the line AB take any point, as C. Join D and C by the straight line D C. E D F B C From the point D draw the straight line D E, making with the line DC an 2, CDE, equal to the DCB. (Prop. XXIII.) Produce the line E D through the point D to any point F. EF is the line required. For since the two straight lines EF and AB are met by a third straight line, and the alternate s so formed, namely, CDE and DC B, are equal, the lines EF and A B are parallel. (Prop. XXVII.) PROPOSITION XXXII. If a side of a triangle is produced, the exterior angle is equal to the sum of the two interior and opposite angles; and the three interior angles of a triangle together equal to two right angles. are For the construction employed in this proposition we must be able to draw through a given point a right line parallel to a given right line. (Prop. XXXI.) For proving the proposition by the aid of the con- 1. That if the same, or equal quantities be added s 3. That if two straight lines are parallel and a Let A B C be a ▲, of which one side, A C, has been produced. B E It has to be shown that the exterior BCD is equal to the sum of the two interior and opposite DS, BAC and ABC, and that the three interiors of the A A B C are together equal to two rights. C Through the point C draw the straight line CE parallel to the side A B. The exterior BCD is thus divided into two parts; and the proof of the proposition consists in showing that one of these parts is equal to one of the interior and opposites; and the other part to the other. Since the lines A B and C E are parallel and are met by a third line, B C, it follows (Prop. XXIX.), that the alternates so formed, namely A B C and B C E, are equal. Again, since A B and C E are parallel, and are intersected by a third line, A D, it follows that the exterior ECD is equal to the interior and opposite on the same side of the intersecting line, namely the BA C. But if equals be added to equals, the sums are equal. (Ax. II.) Therefore, if the equals ABC and BCE be added respectively to the s BAC and ECD, the sums will be equal: that is, the sum of the s BCE and E C D will be equal to the sum of the SABC and B A C. BCE and E C D together make up the exterior BC D. Therefore the exterior B C D is equal to the sum of the two interior and opposites, ABC and BAC. Now, if to each of these equals be added the BCA, the sums will be equal. (Ax. II.) That is to say, the sum of the exterior B CD, and the BCA, will be equal to the sum of the s ABC, BA C, and B C A But since the straight line BC meets the straight line A D, the sum of the adjacent s BCD and BCA is equal to two rights. II |