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Book III   Proposition 36
But
    the squares on EB, BD are equal to the square on ED,
for the angle EFC is right; [I.47]
therefore
    the rectangle AD, DC together with the square on EB is equal to the squares on EB, BD.

    Let the square on EB be subtracted form each;
therefore
    the rectangle AD, DC which remains is equal to the square on DB.

Being what it was required to prove
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