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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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