nd throngh C drawCG II to EF. Then ECGF is the required ||.Because As ABE, AEC are on equal bases BE, EC, andbetween the same Ij*^, they are equal (II., 7, Cor.); andAABC is double of AAEC. Also, because AAEC and |p« ECGF are on the samebase EC and between the same ||, the || ECGF is doubleof AAEC.… … Let ABCD l>e tlie given rectilineal figure, and E thegiven Z. It is required to deserihe a ||* equal to ABCD, andhaving an Z equal to ZE. Join DB.Through C draw CF || to DB, to meet AB produced in F.Join DF.Then AFDB=ACDB, (H., 7.)and to each adding A DAB,ADAF = figure ABCD.Describe the W^ GFHL=…
Geometry · Geometry, Plane
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