Kitap 1
42
And, since BE is equal to EC, the triangle ABE is also equal to the triangle AEC, for they are on equal bases BE, EC and in the same parallels BC, AG; [I. 38] therefore the triangle ABC is double of the triangle AEC.
42
But the parallelogram FECG is also double of the triangle AEC, for it has the same base with it and is in the same parallels with it; [I. 41] therefore the parallelogram FECG is equal to the triangle ABC.
42
And it has the angle CEF equal to the given angle D.
42
Therefore the parallelogram FECG has been constructed equal to the given triangle ABC, in the angle CEF which is equal to D. Q. E. F.
Proposition 43.
43
Enunciation In any parallelogram the complements of the parallelograms about the diameter are equal to one another.
43
Proof. Let ABCD be a parallelogram, and AC its diameter; and about AC let EH, FG be parallelograms, and BK, KD the so-called complements;
43
I say that the complement BK is equal to the complement KD.
43
For, since ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ACD. [I. 34]
43
Again, since EH is a parallelogram, and AK is its diameter, the triangle AEK is equal to the triangle AHK. For the same reason the triangle KFC is also equal to KGC.
43
Now, since the triangle AEK is equal to the triangle AHK, and KFC to KGC, the triangle AEK together with KGC is equal to the triangle AHK together with KFC. [C.N. 2]
43
And the whole triangle ABC is also equal to the whole ADC; therefore the complement BK which remains is equal to the complement KD which remains. [C.N. 3]
43
Therefore etc.
43
Q. E. D.
Proposition 44.
44
Enunciation To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle.
44
Proof. Let AB be the given straight line, C the given triangle and D the given rectilineal angle; thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C.
