T H E E L E M E N T S |
Book VI Proposition 20 | |
Let BE, EC, GL, LH be joined. Now, since the polygon ABCDE is similar to the polygon FGHKL, [1.14] the angle BAE is equal to the angle GFL; and, as BA is to AE, so is GF to FL. Since then ABE, FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, therefore the triangle ABE is equiangular with the triangle FGL; so that it is also similar; therefore the angle ABE is equal to the angle FGL. |
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