PROBLEM K. Let AB be the given straight line. Through A draw AC 1 to AB, from AC cut off AH = AB. For, by construction, AK is a 0. (1. 26) and AH=BK, (1. 26) also AH is = AB; .. AK is equilateral; also :;: BAH is a right 2; ... all the zs of A K are right Zs; (1. 27) AK is both equilateral and rectangular, and is . . a square. .. the AREAS. PROPOSITION XXXI. Parallelograms on the same base and between the same parallels are equal. A E D F в Let the Os ABCD, EBCF be on the same base BC, and between the same parallels AF, BC. Then shall the O ABCD be = the O EBCF. If the sides opposite the base terminate in the same point D, then each of the Os is double of the a DBC; (1. 26) .. O ABCD is = 0 DBCF. DEFINITIONS. The greatest perpendicular which can be let fall from a point in the boundary of a figure on the base or base produced is called the altitude of the figure. The perpendicular let fall on the base or base produced of a s from the vertex is called the altitude of the a. Perpendiculars let fall on the base or base produced of a o from points in the opposite side are equal to one another, and any one of them is spoken of as the altitude of the o. If any one of the sides of a rectangle be taken as the base, then either of the adjacent sides is equal to the altitude. is rectangles, whose bases and altitudes are equal, are themselves equal in all respects. (1. 29) COROLLARIES TO PROPOSITION XXXI. A D is = the rectangle on the same base and between the same parallels. i.e. a o is = the rectangle having the same base and altitude; but all rectangles having equal bases and altitudes are equal. .. a O is = any rectangle of equal base and altitude. .. all parallelograms, having equal bases and altitudes, are equal. Thus parallelograms upon equal bases and between the same parallels are equal. PROPOSITION XXXII. If a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram is double of the triangle. WWWWWOMEN Β BC Let the D EBCD and the A ABC be on the same base BC, and between the same parallels. Then shall D EBCD be double of - ABC. ' If the vertex A of the A ABC is at D or E the proposition is evident. (1. 26) But if not through A draw AX || to EB or DC, meeting the base or base produced in X. Cor. A triangle is half of the rectangle on the same base and between the same parallels. i.e. a triangle is half the rectangle having the same base and altitude. .. a triangle is half of any rectangle of equal base and altitude. .. all triangles having equal bases and altitudes are equal. Thus triangles on the same or equal bases and between 'e same parallels are equal. PROPOSITION XXXIII. Equal triangles on the same base and on the same side of it are between the same parallels. Let FGH, PGH be equal a s upon the same base GH, and upon the same side of it. Join FP. For if not through F draw FQ Il to GH, meeting PG or PG produced in Q, and join QH. Then :: FQ is || to GH; .:. A FGH is = A QGH, (1. 32) but 4 FGH is = A PGH; (hyp.) AQGH is = PGH, which is impossible. ... the temporary hypothesis is false, which was that FP is not || to GH. .. FP is || to GH. Cor. By a similar proof it may be shewn that equal triangles upon equal bases in the same straight line and on the same side of that line are between the same parallels. |