8. Indicate a proof of XXVII. by means of XVII. applied in place of XVI. 9. Also of XXX. by the application of the first part of XXVIII. in place of XXVII. 10. Also of XXX. by the application of the second and third parts of XXIX. and of the second part of XXVIII. 11. Show how to solve XXXI., using the first part of XXVIII. in place of XXVII. 3. Infer that the angles B, ADC are equal, quoting an axiom. F 4. Prove that the angles A and BCD are equal. 5. Show that the angles A, B are together equal to two right angles. (Apply third part of XXIX.) 6. Prove that the angles BCD, CDA are together equal to two right angles. 7. Show that all the angles of a parallelogram are equal to four right angles.. The remaining questions refer to the triangle ABC, which has BA produced to F, and DAE drawn parallel to BC. 8. What is the sum of the angles DAB, BAC, CAE? 9. Show that the same angles are respec- D tively equal to B, BAC, and C of the triangle. 10. Infer that the angles of the triangle ABC are together equal to two right angles. 11. Which angle of the figure is an exterior angle of the triangle? B F E 12. State the two angles of which CAF is the sum? Which of these is equal to B, and which to C? 13. Show that the whole angle CAF is equal to B and C together. 14. Add CAB to each of these equals, and from the result draw an inference with regard to the sum of the angles of the triangle ABC. 75. PROPOSITION XXXII.—THEOREM. If a side of a triangle be produced, the exterior angle is equal to the two opposite interior angles; and the interior angles of every triangle are together equal to tivo right angles. Let ABC be a triangle, having one side BC produced to D. Then the exterior angle ACD shall be equal to the two opposite interior angles CAB, ABC; and the three interior angles ABC, BCA, CAB shall be equal to two right angles. Through C draw CE parallel to the side BA. Then, because AC meets the parallels CE, BA, (I. 31) the angle ACE is equal to the alternate angle BAC. (I. 29) Also, because BC falls upon the parallels CE, BA, therefore the exterior angle ECD is equal to the opposite in terior angle ABC. But the angle ACE was shown to be equal to BAC; (I. 29) therefore the whole exterior angle ACD is equal to the two opposite interior angles CAB, ABC. To each of these equals add the angle ACB; (Ax. 2) then the angles ACD, ACB are equal to the three angles CAB, ABC, and ACB; but ACD, ACB are equal to two right angles; (Ax. 2) (I. 13) therefore, also, the angles CAB, ABC, and ACB are equal to two right angles. Wherefore, if a side of a triangle be produced, &c. EXERCISE XXXVII. (Ax. 1) Q. E. D. 1. If a triangle be equilateral, determine the magnitude of each angle in terms of a right angle. 2. If a triangle be both right-angled and isosceles, determine the value of each acute angle. 3. If a right-angled triangle have one of its acute angles one-third of a right angle, determine the other one. 4. If, in the triangles ABC, DEF, the angles A, D are equal, and also B, E; show that C is equal to F. 5. If an exterior angle of a triangle be three-fourths of a right angle, and one of the opposite interior angles is one-fourth of a right angle; determine the other opposite interior angle. 6. In the figure to Corollary 1 (next page), show that all the angles of the triangles ABF, FBC are together equal to the angles of the quadrilateral ABCF. 7. Show that the sum of the angles of ABCF is equal to four right angles. 8. The given five-sided figure is divided into three triangles; how many right angles are there in the sum of the angles of those three triangles ? 9. Show that the polygon ABCDE has all its angles together equal to all the angles of the three triangles ABC, ACD, ADE. 10. Show that the sum of the angles of the polygon ABCDE is equal to six right angles. B E 11. Make a six-sided figure ABCDEF, and divide it into triangles by drawing straight lines from one angle to each of the others. How many triangles are thus formed, and what is the sum of all their angles? 12. Show that the sum of the angles of a six-sided figure is equal to eight right angles. 13. Divide a five-sided figure into triangles by drawing straight lines from each angle to a point within the figure, as in the figure of Corollary 1, which follows. How many triangles do you thus obtain, and what is the sum of all their angles? 14. Say what is the sum of the angles around the vertex F, giving reference. 15. Considering F the common vertex of all the triangles, deduce from 13 and 14 above the sum of all the angles at the bases of those triangles. 16. Compare the sum of the angles of the polygon, and the sum of the base-angles referred to in 15. Infer the sum of the angles of the polygon in right angles. 76. SIMSON'S COROLLARIES TO PROPOSITION XXXII. The Corollaries which follow might fairly be styled 'propositions," having regard either to their importance or their degree of difficulty. The difficulty which the learner may experience is, however, like that of the Fourth Proposition, rather a matter of words and phraseology than of ideas. A special effort should therefore be made to gradually acquire command of the words; taking care, of course, to follow their meaning at the same time; and the difficulty will then be much lessened. 77. COROLLARY 1.-THEOREM. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. 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 it to each angle. E F B Then, because the three angles of a triangle are equal to two right angles, and there are as many triangles as the figure has sides, therefore all the angles of these triangles are equal to twice as many right angles as the figure has sides. But the interior angles of the figure, together with the angles at the point F, are equal to all the angles of these triangles ; and the angles at F are equal to four right angles; (I. 15, Cor. 2) therefore the angles of the figure, together with four right angles, are equal to the angles of all the triangles. But it has been proved that the angles of the triangles are equal to twice as many right angles as the figure has sides. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides. Q. E. D. EXERCISE XXXVIII. 1. If a figure be a pentagon, or five-sided, how many right angles make twice as many as the figure has sides? What must be added to the sum of the angles of the pentagon in order to make it equal to this number of right angles? 2. How many right angles are therefore equal to the sum of the angles of a pentagon? 3. If a figure be a hexagon, or six-sided, what number of right angles are equal to its six angles and four right angles together? 4. Deduce the sum of the angles of a hexagon. 5. Similarly obtain the sum of the angles of any quadrilateral. 6. Show that the second part of Prop. XXXII. is a particular case of Corollary I. 7. If a quadrilateral have three angles right, what will be the magnitude of the fourth? 8. What proposition is equivalent to this rule ?— To find the sum of the angles of any rectilineal figure, take twice as many right angles as the figure has sides, and subtract four right angles. 9. Apply this rule to find the sum of the angles of a hexagon. From the result deduce the magnitude of each angle when the hexagon is equiangular. 10. Similarly find the magnitude of each angle of an equiangular pentagon, octagon, decagon, duodecagon, and quindecagon respectively; of which the last four have eight, ten, twelve, and fifteen sides respectively. |