Proposition 47.
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.
Let ABC be a right-angled triangle having the angle
BAC right;
I say that the square on BC is equal to the squares on BA, AC.
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.
Then, since each of the angles BAC, BAG is right, it follows that with a straight line BA, and at the point A on it, the two straight lines
AC, AG not lying on the same side make the adjacent angles equal to two right angles;
For the same reason
And, since the angle DBC is equal to the angle FBA: for each is right: let the angle ABC be added to each;
And, since DB is equal to BC, and FB to BA, the two sides AB, BD are equal to the two sides FB, BC respectively,
Now the parallelogram BL is double of the triangle ABD, for they have the same base BD and are in the same parallels
BD, AL. [I. 41]
And the square GB is double of the triangle FBC, for they again have the same base FB and are in the same parallels FB, GC. [I. 41]
[But the doubles of equals are equal to one another.]
Similarly, if AE, BK be joined, the parallelogram CL can also be proved equal to the square HC;
And the square BDEC is described on BC,
Therefore the square on the side BC is equal to the
squares on the sides BA, AC.
Therefore etc.
Q. E. D.