PROPOSITION XXXI. PROBLEM. To draw a straight line through a given point parallel to a given straight line. Let A be the given pt., BC the given st. line. It is required to draw through the pt. A a st. line || BC. F A E Constr. In BC take any pt. D, join AD, and at the pt. A make _DAE = LADC; (i. 23) and produce EA to F. Then shall EF be || BC. Dem. : AD, which meets EF, BC, makes alt. Ls EAD, ADC equal to one another, (constr.) .. EF is || BC. (i. 27) Therefore, through the given pt. A, has been drawn a st. line EAF || the given st. line BC. Q. E. F. 1. Given a point, an angle, and a straight line. From the point draw a line to the given line making with it an angle equal to the given angle. 2. Draw a line DE parallel to the base BC of a triangle ABC, so that DE shall be equal to the sum of BD, CE . 3. Draw a line DE parallel to the base BC of a triangle ABC, so that DE shall be equal to the difference of BD, CE. 4. Through a given point draw a line so that the part of it intercepted between two given parallel lines may be equal to a given straight line. Show that there may be two, one, or no solutions. 5. Through two given points draw two lines forming with a line given in position an equilateral triangle. PROPOSITION XXXII. THEOREM. If a side of any triangle be produced, the exterior angle is equal to the two interior and opposite angles; and the three interior angles of every triangle are equal to two right angles. Let ABC be a A, and let one of its sides BC be produced to D. Then the ext. LACD shall = the two int. and opposite Zs CAB, ABC: and the three int. Ls ABC, BCA, CAB shall together = two rt. Zs. B Constr. Through C draw CE || BA. (i. 31) Dem. Then : CE is || BA and AC meets them, .. LACE = alt. Z BAC. (i. 29) Again :: CE || AB, and BD falls on them, .. ext. LECD = int. and opp. _ ABC; (i. 29) but LACE was shown < BAC; . whole ext. LACD= two intand opp. Ls CAB, ABC: (ax. 2) to each of these equals add LACE, :: LS ACD and ACB = the three s CAB, ABC, ACB; (ax. 2) but LS ACD, ACB together = two rt. ZS, (i. 13) :: Ls CAB, ABC, ACB together = two rt. Zs. (ax. 1) Therefore, if a side of any triangle, &c. Q. E. D. Cor. 1. 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 figure ABCDE can be divided into as many As as it has sides, by drawing st. lines from a pt. F within it to each L. B Now the three int. Ls of each of these As=two rt. Ls. .. all the s of all these As=twice as many rt. Ls as the figure has sides. But the same <s of these as = all the int. As of the figure together with Ls at F; and the Ls at F=four rt. Ls (i. 15. Cor. 2) .. the same Ls of these As=all the int. Ls of the figure together with four rt. Ls. But it has been proved (*) that the Zs of the As are = to twice as many rt. _ s as the figure has sides ; .. all the int. Ls of the figure together with four rt. <s= twice as many rt. Ls as the figure has sides. Cor. 2. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. B Since every int. L such as ABC, with its adj. ext. LABD = two rt. L's, (i. 13) .. all the int. Ls, together with all the ext. Ls,=twice as many rt. Zs as the figure has sides ; but by Cor. 1, all the int. Ls together with four rt. Ls= twice as many rt. Zs as the figure has sides; .. all the int. Zs together with all the ext. Ls are equal to all the int. Ls and four rt. Ls; take away from these equals all the int. ZS, ... all the ext. Ls of the figure are equal to four rt. Zs. [From this prop. it follows : (1) that each of the equal angles of an isosceles right-angled triangle is half a right angle ; (2) that each of the three angles of an equilateral triangle is two-thirds of a right angle; (3) when two angles of a triangle are together equal to the third, the latter is a right angle.] 1. On a given line describe a square of which the given line shall be a diagonal. 2. The difference of the angles at the base of any triangle is double the angle contained by two lines drawn from the vertex, one bisecting the vertical angle, the other perpendicular to the base. 3. Straight lines bisecting two adjacent angles of a parallelogram are at right angles to one another. 4. If the interior angle at one angular point of a triangle and the exterior angle at another be bisected by straight lines, the angle contained by the two bisecting lines is equal to half the third angle of the triangle. 5. ABC is a triangle right-angled at A with the angle B double of the angle C; show that CB is double of AB. 6. In the figure to i. 5, if BG, FC intersect at H, and if the angle FBG is equal to the angle ABC, then the angle BHF is equal to twice the angle BAC. 7. To construct a triangle when one angle, a side opposite to it, and the sum of the other two sides are given. 8. What figure is that, whose exterior angles (formed by producing the sides successively) are together equal to its interior angles. 9. How many sides has a figure, if the sum of its interior is double that of its exterior angles ? 10. ABC is an equilateral triangle; D, E are points in BC, CA respectively such that BD is equal to CE; if AD, BE be joined and intersect in O, show that the angle AOB is twice an angle of the equilateral triangle. 11. If each of the equal angles of an isosceles triangle be one-fourth the vertical angle, and from one of them a perpendicular is drawn to the base meeting the opposite side produced; then will the part produced, the perpendicular, and the remaining side form an equilateral triangle. |