T H E E L E M E N T S |
Book III Proposition 29 | |
Now, since the circumference BGC is equal to the circumference EHF, the angle BKC is also equal to the angle ELF. [III.27] And, since the circles ABC, DEF are equal, the two sides BK, KC are equal to the two sides EL, LF; and they contain equal angles; therefore the base BC is equal to the base EF. [I.4] Being what it was required to prove. |
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