Items related to Euclid for beginners, books i. and ii., with simple...
Synopsis
This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1880 Excerpt: ...ABCD, and having an angle = the angle E. Construction.--1. Join BD and describe the parallelogram FH = the triangle ABD, and having the angle FKH = the angle E (I. 42). 2. Apply to the straight line GH the parallelogram GM equal to the triangle DBC, and having the angle GHM = the angle E (I. 44). Then it is to be proved that FM is a parallelogram = the figure ABCD, and having the angle FKM = the angle E. Proof.--Because each of the angles FKH and GHM = the angle E (cons.), therefore the angle FKH = the angle GHM (ax. 1); add to each the angle GHK, then the angles FKH and GHK = the angles GHM and GHK (ax. 2). But the angles FKH and GHK = two right angles (I. 29); there/ore the angles GHM and GHK = two right angles, and therefore KH is in the same straight line with HM (I. 14). 2. Next, because the lines FG and KM are parallel (cons.), and GH meets them, therefore the alternate angle FGH = the alternate angle GHM (I. 29); add to each the angle LGH, then the angles FGH and LGH = the angles GHM and LGH (ax. 2). But the angles MHG and LGH = two right angles (I. 29); therefore the angles HGF and LGH = two right angles, and therefore FG is in the same right line with GL (I. 14). 3. Next, hecause KF is parallel to HG, and HG to ML (cons.), therefore KF is parallel to ML (I. 30), and. KM and FL are also parallel (cons.); therefore the figure FKML is a parallelogram (def. 34). Again, because the parallelogram FH = the triangle ABD (cons.), and the parallelogram GM = the triangle DBC (cons.), therefore the parallelogram FM = the figure ABCD, and it has the angle FKM = the angle E (cons.) Therefore, it is proved, as required, that FM is a parallelogram = the figure ABCD, and having an angle FKM = the angle E. Q. E. F. Corollary.--A parallelogram may be described equal ...
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