An elementary treatise on geometry : simplified for beginners not versed in algebra . equal parts. I. Solution. Let AB be thegiven line, and let it be required todivide it into five equal parts. 1. From the point A, draw anindefinite straight line AN, makingany angle you please with the lineAB. 2. Take any distance Am, andmeasure it off 5 times upon the lineAN. 3. Join the last point of division q,and the extremity B of the line AB. 4. Through m, ?z, o, p, q, draw the straight limes bm,en, do, ep, parallel to Bq; the line AB is divided into fiveequal parts. The demonstration follows immediatel
An elementary treatise on geometry : simplified for beginners not versed in algebra . equal parts. I. Solution. Let AB be thegiven line, and let it be required todivide it into five equal parts. 1. From the point A, draw anindefinite straight line AN, makingany angle you please with the lineAB. 2. Take any distance Am, andmeasure it off 5 times upon the lineAN. 3. Join the last point of division q,and the extremity B of the line AB. 4. Through m, ?z, o, p, q, draw the straight limes bm,en, do, ep, parallel to Bq; the line AB is divided into fiveequal parts. The demonstration follows immediately from Query 14, Sect. 11. II. Solution. Let AB be thegiven straight line, which is to bedivided into 5 equal parts. 1. Draw a straight line MN,greater than AB, parallel to AB. 2. Take any distance Mn, andmeasure it off 5 times upon theline MN. 3. Join the extremities of boththe lines Mr and AB, by the straightlines MA, rB, which will cut eachother, when suflEiciently extended,in a point I. 4. Join In, lo, Ip, Iq, the line AB is divided into 5equal parts, viz. A6, be, cd, de, GEOMETRY. 149 Demox. The triangles A6I, bcl, cdl, del, cBI, are similar lothe triangles Mril, nol, opi, j?^!, ^rl, each to each; because theline AB is drdiwn parallel to Mr (Query 16, Sect. II.) ; and as thebases Mn, no, op,pq, qr, of the latter triangles are all equal to oneanother, the bases A6, he, cd, de, eB of the former triangles mustalso be equal to one another. Remark. If it were required to divide a line into two partswhich shall be in a given ratio, for instance, as 2 to 3, you needonly, as before, take 5 equal distances upon the line MN, and thenjoin the point I to the second and last point of division ; the lineAB will, in the point c, be divided in the ratio of 2 to 3. In a simi-lar manner can any given straight line be divided into 3, 4, 5, &, which shall be to each other in a given ratio. Problem X. Three lines being given, to find a fourthone, which shall he in a geometrical pro
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