T H E E L E M E N T S |
Book VI Proposition 25 | |
Let there be applied to BC the parallelogram BE equal to the triangle ABC, [I.44] and to CE the parallelogram CM equal to D and in the angle FCE which is equal to the angle CBL. [I.45] Therefore BC is in a straight line with CF, and LE with EM. Now let GH be taken a mean proportional to BC, CF [VI.13], and on GH let KGH be described similar and similarly situated to ABC. |
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