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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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