T H E E L E M E N T S |
Book I Proposition 45 | |
And, since the triangle ABD is equal to the parallelogram FH, and DBC to GM, the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. Therefore, the parallelogram KFLM has been constructed equal to the given rectilineal figure ABCD, in the angle FKM which is equal to the given angle E. Being what it was required to do. |
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