. Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson. and for similar reason the parallel-ograms HBAG and MBDA; hence (I. ax. 1) MBDA is equal toBLHW. In similar manner, it can be shown tlnit CEFA isequiA^alent to LCKW; therefore the whole parallelogram BCHK(I, axs. 2 and 10) is equivalent to the sum of the two parallelo-grams MB AD and CEFA. Wherefore, if on any two sides ofa
. Geometry : the elements of Euclid and Legendre simplified and arranged to exclude from geomtrical reasoning the reductio ad absurdum : with the elements of plane and spherical trigonometry, and exercises in elementary geometry and trigonometry / /c By Lawrence S. Benson. and for similar reason the parallel-ograms HBAG and MBDA; hence (I. ax. 1) MBDA is equal toBLHW. In similar manner, it can be shown tlnit CEFA isequiA^alent to LCKW; therefore the whole parallelogram BCHK(I, axs. 2 and 10) is equivalent to the sum of the two parallelo-grams MB AD and CEFA. Wherefore, if on any two sides ofa triangle, etc. Cor. 1. A paiticular case of this proposition is, lohen the tri-angle is right-angled., then the sqiuires described on the legs—that is, the sides containing the right angle, are together equiva-lent to the square on the liypotlieniise—that is, the side oppositethe right angle. Let ABC be a rio-ht-anoledtriangle, having the right angleBAC; the square described ontlie hypothenuse BC is equiva-lent to the sum of the squareson BA and AC. On BC, BA,Q and AC (L 23) desciibe thesquares BCHK, BADE, andACFG; through A draw ALparallel to BH (L 18), anddraw AH and DC (L post. 1).Tlien, because the anglesBAC and BAE are both rightangles (L def 10), the two. BOOK I.] EUCLID AND LEGENDEE. 43 straight lines CA and AE are in the same straight line (I, 10).For like reason, BA and AF are in the same straight : because the angle IIBC is equal to the angle DBA (I,ax. 1), if the angle ABC be added to each, we have (I. ax. 2)HBA equal to DBC ; and because AB is equal to DB (const.),and BIT equal to BC, therefore (I. 3, case 1) the triangle ABHis equal to the triangle DBC. But (I. 15, cor. 4) the parallelo-gram BPIIL is double the ti-iangle ABH; for like reason thesquare BADE is double the triangle DBC ; hence (I. ax. 6)BPIIL is equivalent to BADE. In like manner, PCLK can beshown equivalent to ACFE. Xow (I. ax. 10), BCHK is equiva-lent to BPHL and PCLK together; hence
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