T H E E L E M E N T S |
Book IV Proposition 5 | |
Again, let DF, EF meet outside the triangle ABC at F,
as is the case in the third figure, and let AF, BF, CF be joined. Then, again, since AD is equal to BD, and DF is common and at right angles, therefore the base AF is equal to the base BF. [I.4] Similarly we can prove that CF is also equal to AF; so that BF is also equal to CF; |
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