7m ABCD = EFGH. Ax. 1. Wherefore parallelograms, &c. COR. 1.-Triangles upon the equal bases and between the same parallels are equal to each other. For draw the diams. AC, EG; the As ABC, EFG, are the halves of the equal ms ☐m3 ABCD, EFGH, and are therefore equal Ax. 7. to each other. COR. 2.-If a parallelogram and a triangle be upon the same or equal bases, and between the same parallels, the parallelogram is double the triangle. For the ABCD is double the ABC, or double the is its equal, by last Cor. EFG, which PROP. XXXVI. THEOR. 39. 1 Eu. Equal triangles upon the same base, and upon the same side of it, are between the same parallels. Let the As ABC, DBC, be on the same base BC, and on the same side of it; then the As will be between the same ||s. In the same manner it may be proved that no other line but AD || BC, AD || BC. Wherefore equal triangles upon, &c. COR.--In the same way it may be shown, that equal triangles upon equal bases and towards the same parts, are between the same parallels PROP. XXXVII. THEOR. 43. 1 Eu. The complements of the parallelograms which are about the diameter of any parallelogram, are equal to each other." Let ABCD be a 7m, AC its diam.; and 7ms about AC, that is, passes. And BEKG, 'which make up the ms whole figure ABCD, are called complements. Then comp. BEKG = comp. KHDF. { ABC = A ADC; ABCD, Prop.33. ΔΑΕΚ = Δ ΑΗΚ, .*; ^ AEK +◇ KGC ={◇ AHK + ◇AX. 2. but the whole .. KFC: ABC whole ADC; remains comp. the remains comp. BEKG ·}={ KHDF. Wherefore the complements, &c. Ax. 3. PROP. XXXVIII. PROB. 46. 1 Eu. To describe a square upon a given straight line. Let AB be the given str. line; it is required to descr. a square upon AB. Also Prop. 28. Constr. Ax. 3. Prop. 33. JADEB is TM, and its opp. sides are equal; AB = = AD=DE= EB, and ADEB is equilat. AD meets the ||s AB, DE, :: <BAD+ADE but = 2 rt. s but opp.s of ms are equal, Ax. 1. thes ABE, BED, are rt. Ls, .. the figure ADEB is rectangular and equilat. Def. 30... the figure is a square, and descr. on AB. COR.-Every that has one rt., has all its angles rt. s. PROP. XXXIX. THEOR. 47.1 Eu. In any right-angled triangle, the square which Let ABC be a rt. angled, having the rt. = On BC, BA, AC, descr. the squares BE, Prop. 38. GB, HC.* Draw AL || BD or CE; Prop. 30. join AD, FC. BAG = 2 rt. s. Hyp. & Def. 30. CA is in the same str. line with AG. Prop. 13. <BAC+= For the same reason AB is in the same str. line with AH, and DBC = / FBA, ea. being a rt. • Def. 30. Add to each / ABC; Squares and parallelograms are frequently expressed, for the sake of brevity, by the letters at their opposite angles; and the square on any line, as BC, is represented by BC2. |