PROPOSITION 14. THEOREM. If, at a point in a straight line, two other straight lines, on opposite sides of it, make the adjacent angles together equal to two right angles, then these two straight lines shall be in one and the same straight line. A B At the point B in the straight line AB, let the two straight lines BC, BD, on the opposite sides of AB, make the adjacent angles ABC, ABD together equal to two right angles. Then BD shall be in the same straight line with BC. Proof. For if BD be not in the same straight line with BC, if possible, let BE be in the same straight line with BC. Then because AB meets the straight line CBE, therefore the adjacent angles CBA, ABE are together equal to two right angles. I. 13. But the angles CBA, ABD are also together equal to two right angles. Hyp. Therefore the angles CBA, ABE are together equal to the angles CBA, ABD. Ax. 11. From each of these equals take the common angle CBA; then the remaining angle ABE is equal to the remaining angle ABD; the part equal to the whole; which is impossible. Therefore BE is not in the same straight line with BC. And in the same way it may be shewn that no other line but BD can be in the same straight line with BC. Therefore BD is in the same straight line with BC. Q.E.D. EXERCISE. ABCD is a rhombus; and the diagonal AC is bisected at O. If O is joined to the angular points B and D; shew that OB and OD are in one straight line. PROPOSITION 15. THEOREM. If two straight lines intersect one another, then the vertically opposite angles shall be equal. B A Let the two straight lines AB, CD cut one another at the point E. Then (i) the angle AEC shall be equal to the angle DEB; Proof. Because AE meets the straight line CD, therefore the adjacent angles CEA, AED are together equal to two right angles. I. 13. Again, because DE meets the straight line AB, therefore the adjacent angles AED, DEB are together equal to two right angles. I. 13. Therefore the angles CEA, AED are together equal to the angles AED, DEB. From each of these equals take the common angle AED; then the remaining angle CEA is equal to the remaining angle DEB. Ax. 3. In the same way it may be proved that the angle CEB is equal to the angle AED. Q.E.D. COROLLARY 1. From this it follows that, if two straight lines cut one another, the four angles so formed are together equal to four right angles. COROLLARY 2. Consequently, when any number of straight lines meet at a point, the sum of the angles made by consecutive lines is equal to four right angles. If one side of a triangle be produced, then the exterior angle shall be greater than either of the interior opposite angles. Let ABC be a triangle, and let BC be produced to D. Then shall the exterior angle ACD be greater than either of the interior opposite angles ABC, BAC. Construction. Bisect AC at E; I. 10. Join BE; and produce it to F, making EF equal to BE. I. 3. Join FC. Proof. Then in the triangles AEB, CEF, and EB is equal to EF; Constr. Constr. Because also the angle AEB is equal to the vertically opposite angle CEF; I. 15. therefore the triangle AEB is equal to the triangle CEF in all respects: I. 4. so that the angle BAE is equal to the angle ECF. But the angle ECD is greater than its part, the angle ECF; therefore the angle ECD is greater than the angle BAE; that is, the angle ACD is greater than the angle BAC. In the same way, if BC be bisected, and the side AC produced to G, it may be proved that the angle BCG is greater than the angle ABC. But the angle BCG is equal to the angle ACD: 1. 15. therefore also the angle ACD is greater than the angle ABC. Q.E.D. Any two angles of a triangle are together less than two right angles. Let ABC be a triangle. Then shall any two of the angles of the triangle ABC be together less than two right angles. Construction. Produce the side BC to D. Proof. Then because BC, a side of the triangle ABC, is produced to D; therefore the exterior angle ACD is greater than the interior opposite angle ABC. To each of these add the angle ACB: I. 16. then the angles ACD, ACB are together greater than the angles ABC, ACB. Ax. 4. But the adjacent angles ACD, ACB are together equal to two right angles. I. 13. Therefore the angles ABC, ACB are together less than two right angles. Similarly it may be shewn that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles. Q.E.D. NOTE. It follows from this Proposition that every triangle must have at least two acute angles: for if one angle is obtuse, or a right angle, each of the other angles must be less than a right angle. EXERCISES. 1. Enunciate this Proposition so as to shew that it is the converse of Axiom 12. 2. If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles. 3. Shew how a proof of Proposition 17 may be obtained by joining each vertex in turn to any point in the opposite side. If one side of a triangle be greater than another, then the angle opposite to the greater side shall be greater than the angle opposite to the less. D B Let ABC be a triangle, in which the side AC is greater than the side AB. Then shall the angle ABC be greater than the angle ACB. Construction. From AC cut off a part AD equal to AB. Proof. Join BD. Then in the triangle ABD, because AB is equal to AD, therefore the angle ABD is equal to the angle ADB. I. 3. I. 5. But the exterior angle ADB of the triangle DCB is greater than the interior opposite angle DCB, that is, greater than the angle ACB. I. 16. Therefore also the angle ABD is greater than the angle ACB; still more then is the angle ABC greater than the angle ACB. Euclid enunciated Proposition 18 as follows: Q.E.D. The greater side of every triangle has the greater angle opposite to it. [This form of enunciation is found to be a common source of diffi culty with beginners, who fail to distinguish what is assumed in it and what is to be proved. If Euclid's enunciations of Props. 18 and 19 are adopted, it is important to remember that in each case the part of the triangle first named points out the hypothesis.] |