But EDC (Const.), and therefore the given line AB is so divided that the rectangle AG GB contained by its segments is equal to the square of C. PROP. XXIX. PROB. To produce a given straight line (AB), so that the rectangle contained by the segments between the extremities of the given line and the point to which it is produced, may be equal to a given area. E Bisect AB in D, and from one of the extremities of AB, draw BE at right angles to AB, and make it equal to C, the square of which is equal to the given area; join DE; and from the centre D, with the interval DE, describe a circle meeting AB produced to G. Then AG GB+DB' =DG2 (ii. Prop. 6) DE2. But DE2-EB2+DB2 (i. Prop. 47); therefore AG GB + DB2 = EB2 + DB2, and DB2 being taken from both, AG⚫GB = EB2; but EB is equal to C, the square of which is equal to the given area; EB2 is therefore equal to that area, and consequently the given line AB has been produced to G in such a manner that the rectangle contained by the whole produced line and the produced part is equal to the given area. PROP. XXX. PROB. To divide a given straight line (AB) in extreme and mean ratio. E On AB describe the square AD, and produce its side CA to F, so that CF FA = AB2 (vi. Prop. 29); take AG equal to AF, and through G draw KH parallel to CA, and meeting FH drawn parallel to AB. Then the right- A angled parallelogram CH is manifestly equal to the rectangle CF FA, and is therefore equal to square of AB; and if from both these equals the C F H G B be taken the common part AK, the rectangle FG, which is the square of AG, will be equal to GD, which is equal to the rectangle AB GB; therefore AB: AG:: AG: GB, and the given line is divided in extreme and mean ratio (vi. Def. 3). PROP. XXXI. THEOR. In a right-angled triangle (ACB), the rectilinear figure described upon the side (AB) opposite to the right angle, is equal to the similar and similarly described figures upon the sides (AC, CB) containing the right angle. A B From the vertex of the right angle draw CD perpendicular to the opposite side, and it divides the triangle ACB into parts similar to the whole (vi. Prop. 8), so that AB: AC::AC: AD; but the figure on AB is to the similar and similarly placed figure on AC in the duplicate ratio of their homologous sides (vi. Prop. 20), that is, as AB to AĎ: and, in like manner, the figure on AB is to the similar and similarly placed figure on BC, as AB to DB; therefore the figure on AB is to the figures on AC and BC together in the same ratio as AB is to AC and BC together, that is, in a ratio of equality. If two triangles (ABC, DCE), having two sides of the one proportional to two sides of the other (AB: AC :: DC: DE), be joined at one angle, so that their homologous sides be parallel to each other, the remaining sides (BC, CE) shall be in a straight line. Since the side DC is parallel to AB, / BAC=/ ACD (i. Prop. 29); in like manner, and meets AC, ACD=/ CDE; therefore BAC = CDE; and since the sides about these (vi. Prop. 6), and ▲ ABC= DCE: but it has BOL together, equal to two right angles (i. Prop. 32), and consequently, BC and CE lie in the same straight line (i. Prop. 14). E PROP. XXXIII. THEOR. In equal circles (BAL, EDN), angles, whether at the centre or circumference, are in the same ratio to one another as the arches on which they stand ( BGC : EHF:: BC: EF): so also are sectors. D M N Take, in the circumference of the circle BAL, any number of arches CK, KL, each equal to BC, and also in the circumference of the circle DEN any number of arches FM, MN, each equal to EF, and join GK, GL, HM, HN. Then, since the arches BC, CK, KL are all equal, the Zs BGC, CGK, KGL are also equal (iii. Prop. 27); therefore BGL is the same multiple of BGC, which the arch BL is of the arch BC; and, in like manner, EHN is the same multiple of EHF which the arch EN is of the arch EF. But if the arch BL=EN, then also BGL=EHN (iii. Prop. 27); if BLEN, then/BGL/EHN; if BLEN, BGL ▼▲EHN; therefore BGC : ▲ EHF :: BC: EF (v. Def. 5), or the angles at the centre are as the arches on which they stand; but the angles BAC, EDF at the circumference, being halves of the angles at the centre (iii. Prop. 20), are in the same ratio as these (v. Prop. 15), and therefore are also as the arches on which they stand. A E F K N M And the sector BGC also is to the sector EHF as the arch BC to the arch EF. For join BC, CK, and in the arches BC, CK take any points X, O, and join BX, XC, CO, OK. The triangles BGC, CGK, having equal sides containing equal angles, are equal (i. Prop. 4), B and their bases, BC and CK, are equal. But since the arches BC and CK are equal, if they be taken from the whole circumference, the respective remainders will be equal; but these are the arches on which the Zs BXC, COK stand, which are therefore equal (iii. Prop. 27): consequently, the segments BXC, COK are similar (iii. Def. 9), and, as they stand on equal straight lines, BC, CK, they are also equal (iii. Prop. 24) ; and if they be added to the equal triangles BGC, CGK, the sectors BGC, CGK will be equal. In like manner, the sector KGL may be shown to be equal to BGC or CGK, and also the sectors EHF, FHM, MHN may be proved to be equal to one another., Therefore, the sector BGL is the same multiple of the sector BGC which the arch BL is of the arch BC and the sector EHN is the same multiple of the sector EHF, which the arch EN is of the arch EF: and if BL=EN, the sector BGL=EHN; if BLEN, BGL7 EHN; if BLEN, BGLEHN; and BL and EN are any multiples whatever of BC and EF; therefore the sector BGC is to the sector EHF (v. Def. 5), as the arch BC is to the arch EF. |