27 PROPOSITION XV. THEOREM. If two straight lines cut one another, the vertically opposite angles must be equal. Let the st. lines AB, CD cut one another in the pt. E. Then must AEC-L BED and AED= L BEC. For AE meets CD, .. sum of 4s AEC, AED=two rt. 4 s. I. 13. And DE meets AB, .. sum of 48 BED, AED=two rt. 4s; I. 13. Ax. 3. Similarly it may be shewn that AED= 2 BEC. Q. E. D. COROLLARY I. From this it is manifest, that if two straight lines cut one another, the four angles, which they make at the point of intersection, are together equal to four right angles. COROLLARY II. All the angles, made by any number of straight lines meeting in one point, are together equal to four right angles. Ex. 1. Shew that the bisectors of AED and BEC are in the same straight line. Ex. 2. Prove that ▲ AED is equal to the angle between two straight lines drawn at right angles from E to AE and EC, if both lie above CD. Ex. 3. If AB, CD bisect each other in E; shew that the triangles AED, BEC are equal in all respects. NOTE 3. On Euclid's definition of an Angle. Euclid directs us to regard an angle as the inclination of two straight lines to each other, which meet, but are not in the same straight line. Thus he does not recognise the existence of a single`angle equal in magnitude to two right angles. The words printed in italics are omitted as needless, in Def. VIII., p. 3, and that definition may be extended with advantage in the following terms : DEF. Let WQE be a fixed straight line, and QP a line which revolves about the fixed point Q, and which at first coincides with QE. W Then, when QP has reached the position represented in the diagram, we say that it has described the angle EQP. When QP has revolved so far as to coincide with QW, we say that it has described an angle equal to two right angles. Hence we may obtain an easy proof of Prop. XIII.; for whatever the position of PQ may be, the angles which it makes with WE are together equal to two right angles. Again, in Prop. xv. it is evident that AED= ▲ BEC, since each has the same supplementary AEC. We shall shew hereafter, p. 149, how this definition may be extended, so as to embrace angles greater than two right angles. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. A E D Let the side BC of ▲ ABC be produced to D. Then must ACD be greater than either ▲ CAB or ABC. Produce BE to F, making EF=BE, and join FC. Then in AS BEA, FEC, ·· BE=FE, and EA=EC, and ▲ BEA= ▲ FEC, I. 15. I. 4. ECF; Ax. 9. that is, .. ACD is greater than 4 ACD is greater than ▲ CAB. Similarly, if AC be produced to G it may be shewn that and 4 BCG is greater than ABC. L BCGL ACD; .. 4 ACD is greater than ▲ ABC. I. 15. Q. E. D. Ex. 1. From the same point there cannot be drawn more than two equal straight lines to meet a given straight line. Ex. 2. If, from any point, a straight line be drawn to a given straight line making with it an acute and an obtuse angle, and if, from the same point, a perpendicular be drawn to the given line; the perpendicular will fall on the side of the acute angle. Any two angles of a triangle are together less than two right angles. A C Let ABC be any ▲. Then must any two of its s be together less than two rt. 4 s. Then Produce BC to D. 4 ACD is greater than ABC. I. 16. :. ¿s ACD, ACB are together greater than 4s ABC, ACB. But 2 s ACD, ACB together=two rt. ≤ s. I. 13. .. 4s ABC, ACB are together less than two rt. 4 s. Similarly it may be shewn that s ABC, BAC and also that s BAC, ACB are together less than two rt. 4 s. NOTE 4. On the Sixth Postulate. Q. E. D. We learn from Prop. xvII. that if two straight lines BM A, are met by another straight line and CN, which meet in DE in the points O, P, the angles MOP and NPO are together less than two right angles. The Sixth Postulate asserts that if a line DE meeting two other lines BM, CN makes MOP, NPO, the two interior angles on the same side of it, together less than two right angles, BM and CN shall meet if produced on the same side of DE on which are the angles MOP and NPO. PROPOSITION XVIII. THEOREM. one side of a triangle beater than a second, the angle opposite the first must be greater than that opposite the second. B In ▲ ABC, let side AC be greater than AB. Then must ▲ ABC be greater than ▲ ACB. From AC cut off AD=AB, and join BD. I. 3. I. A. And CD, a side of ▲ BDC, is produced to A. A :. 4 ADB is greater than ▲ ACB; .. also ABD is greater than ▲ ACB. Much more is ABC greater than ▲ ACB. I. 16. Q. E. D. Ex. Shew that if two angles of a triangle be equal, the sides which subtend them are equal also (Eucl. I. 6). 215 |