Kitap 1
46
Let AC be drawn at right angles to the straight line AB from the point A on it [I. 11], and let AD be made equal to AB; through the point D let DE be drawn parallel to AB, and through the point B let BE be drawn parallel to AD. [I. 31]
46
Therefore ADEB is a parallelogram; therefore AB is equal to DE, and AD to BE. [I. 34]
46
But AB is equal to AD; therefore the four straight lines BA, AD, DE, EB are equal to one another; therefore the parallelogram ADEB is equilateral.
46
I say next that it is also right-angled.
46
For, since the straight line AD falls upon the parallels AB, DE, the angles BAD, ADE are equal to two right angles. [I. 29]
46
But the angle BAD is right; therefore the angle ADE is also right.
46
And in parallelogrammic areas the opposite sides and angles are equal to one another; [I. 34] therefore each of the opposite angles ABE, BED is also right. Therefore ADEB is right-angled.
46
And it was also proved equilateral.
46
Therefore it is a square; and it is described on the straight line AB.
46
Q. E. F.
Proposition 47.
47
Enunciation In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.
47
Proof. Let ABC be a right-angled triangle having the angle BAC right;
47
I say that the square on BC is equal to the squares on BA, AC.
47
For let there be described on BC the square BDEC, and on BA, AC the squares GB, HC; [I. 46] through A let AL be drawn parallel to either BD or CE, and let AD, FC be joined.
