But the parallelogram ABCD is double of ABC, because its diameter AC bisects it; wherefore ABCD is also double of EBC. (I. 34) Therefore, if a parallelogram, &c. Q. E. D. 91. PROPOSITION XLII.-PROBLEM. To describe a parallelogram that shall be equal to a given triangle and have an angle equal to a given rectilineal angle. Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to ABC, and have an angle equal to D. Bisect BC in E (I. 10), and join AE. At the point E, in the straight line EC, make the angle FEC equal to the angle D. Through A draw AFG parallel to BC, and through C draw CG parallel to EF. Then the figure CEFG is a parallelogram. (I. 23) (I. 31) (Def. A.) Because the triangles ABE, AEC are on equal bases BE, EC, and between the same parallels BC, AG, therefore the parallelogram FECG is equal to ABC; (Ax. 6 and the angle CEF was made equal to D. Wherefore, a parallelogram FECG has been described, equal to the given triangle ABC, and having one angle CEF equal to the given angle D. Q. E. F. 92. ALTITUDES OF TRIANGLES AND PARALLELOGRAMS. As with the triangle, so with the parallelogram, any side may be chosen to be called the base. It will be convenient here, for practical applications of this part of the subject, to define height or altitude of a parallelogram and of a triangle. Definition. The altitude of a parallelogram is a perpendicular drawn to the base, or base produced, from a point in the opposite side, or the opposite side produced. The altitude of a triangle is a similar perpendicular to the base, but is drawn from either the vertex, or from a point in a parallel to the base which passes through the vertex. As examples, these figures ABCD, ABK are a parallelogram and a triangle respectively, and FG, HK, MN are perpendiculars to AB; while, in the second figure, GN is parallel to AB. Then the altitude of either the parallelogram or the triangle may be taken to be any one of the three perpendiculars FG, HK, MN. Wherever this altitude be thus taken, it will be found that it is always of the same magnitude. This, of course, is necessary, and its proof, being easy, will be required from the learner in the ensuing exercise. Another explanation may conveniently be introduced here. When two figures are equal in area, but not necessarily so in other respects, they are sometimes said to be "equivalent." Such figures have come under notice in the present section. EXERCISE XLIX. 1. If from two points in one of a pair of parallels two perpendiculars be drawn to meet the other parallel, show that these perpendiculars shall be equal. 2. Show that the figure obtained, which is bounded by the parallels and the perpendiculars, is a rectangle. 3. Show that, if two parallelograms have the same altitude, they can be placed between the same parallels. 4. Hence show that parallelograms on equal bases and of equal altitudes are themselves equal. 5. Show that a triangle is half of the rectangle upon an equal base and of equal altitude. 6. Make a rectangle equal to a given triangle. 7. Show that a parallelogram is equal to a triangle between the same parallels, but on double its base. 8. Make a parallelogram equal to a given parallelogram, but having one angle at its base equal to half a right angle. 9. Make a rectangle equal to a given parallelogram. 10. Make a triangle equal in area to a given parallelogram. 11. Make a right-angled triangle equal to a given triangle. 12. Make a triangle equal to a given triangle, having an angle at its base equal to a given angle. 13. Show that the parallelograms of Prop. XXXV. are equivalent " figures. 14. Of the following propositions, state which relate to figures "equal in all respects," and which to "equivalent" figures merely: XXXVIII., VIII., XL., IV., XXVI., XXXVI. Note.-If convenient, it would be well to work the Simple Mensuration of the rectangle, parallelogram, and triangle at the same time with this part of Geometry. 93. SECTION X.-PROPOSITIONS XLIII.-XLV. Complementary Parallelograms, and the Reduction of Rectilineal Figures to Parallelograms of equal area. It is curious that in the next proposition two new terms are employed; and that definitions of both are incorporated with the proposition itself. These terms are—" parallelograms about the diameter of a parallelogram," and "complements." A few general hints will here be given to introduce the latter term; and then the student will be left to extract the definitions which suit the proposition from itself. If we take one or more parts of any whole magnitude, that which must be added to those parts to make up the whole, is sometimes called the "complement" of the parts. For example, if we take two-thirds of a square, the remaining one-third might be called the complement of the two-thirds. But it must be specially noticed, that the one-third is not called the complement of the whole square. To take an illustration in numbers, if we have under consideration at any time the number 10 as our whole magnitude, the complement of 3 would be 7; if the whole be unity, the complement of would be ; and so on. EXERCISE L. Throughout this exercise the figure of the next proposition is referred to, in which ABCD, EH, and GF are all parallelograms. 1. Taking ABCD as the whole magnitude, give the complement of the triangle ABC, of the parallelogram BF, and of the parallelogram FG. 2. What will be the complement of the figures BK, KD taken together? 3. Of the parallelograms EH, GF taken together? 4. Of the whole figure ABCD? 5. What proposition proves that the triangles ABC, ADC are equal? Also AEK, AHK? 6. Show that the triangles AEK and KGC are together equal to AHK and KFC. 7. Give the remainders when AEK, KGC are taken from ABC; also AHK, KFC from ADC. 8. By what axiom may these remainders be shown to be equal? 94. PROPOSITION XLIII.-THEOREM. The complements of the parallelograms which are about the diameter of any parallelogram are equal to one another. Let ABCD be a parallelogram, of which the diameter is AC; and EH, GF the parallelograms about AC, that is, which have AC passing through opposite angles in each. Also let BK, KD be the other parallelograms which complete the whole figure ABCD, and which are therefore called the complements. Then the complements BK, KD shall be equal. Because ABCD is a parallelogram, and AC its diameter; therefore the triangle ABC is equal to ADC. (I. 34) Again, because AEKH is a parallelogram, and AK its diameter, therefore the triangle AEK is equal to AHK ; (I. 34) and, for the same reason, the triangle KGC is equal to KFC; wherefore the two triangles AEK, KGC are equal to the two AHK, KFC. (Ax. 2) But the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK is equal to the remaining complement KD. Wherefore, the complements, &c. 95. EXERCISE LI. (Ax. 3) Q. E. D. 1. Write the definition, as given in substance in XLIII., of “parallelograms about the diameter of a parallelogram." 2. Also of "complements" of those parallelograms. 3. Say why it would be wrong to describe BK, KD as the complements of ABCD in the figure of XLIII. 4. Given a straight line AB and a triangle PQR, show how to make a triangle equal to PQR, one side of which shall be in AB produced, and one angle at B equal to a given angle. 5. Show how a parallelogram may now be made equal to PQR having one side in AB produced, and an angle at B equal to a given angle. 6. If one straight line meets two others without cutting them, how many angles are formed? 7. Under what condition will the two lines meet if produced, and in what direction must they be produced to meet? 8. In the figure to XLIV., FL, EM, GA are all parallelograms; show that, if BL be equal to C, then C is equal to BF. 9. Also show that, if the angle ABM is equal to the angle D, then D is equal to the angle EBG. 10. In the figure to XLIII., if the angle BAD be right, show that all the quadrilaterals of the figure are rectangles. 11. In the same figure, if EH be a square, so is the whole BD. Definition. A parallelogram is said to be applied to a straight line, when it is described upon it as one of its sides. 96. PROPOSITION XLIV.-PROBLEM. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Let AB be the given straight line, C the given triangle, and D the given rectilineal angle. |