T H E E L E M E N T S |
Book IV Proposition 15 | |
I say next that it is also equiangular. For, since the circumference FA is equal to the circumference ED, let the circumference ABCD be added to each; therefore the whole circumference FABCD is equal to the whole circumference ABCDE;. and the angle FED stands on the circumference FABCD, and the angle AFE on the circumference ABCDE; therefore the angle AFE is equal to the angle DEF. [III.27] |
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