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
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
is right, if follows that with a straight line
BA, and at the point
A on it, the two straight lines
not lying on the same side make the adjacent angles equal to two right
therefore CA is in a straight line with AG. [I. 14]
For the same reason
BA is also in a straight line with AH.
And, since the angle DBC
is equal to the angle FBA: for
each is right:
let the angle ABC be added to each;
therefore the whole angle DBA is equal to the whole angle FBC. [C.N. 2]
And, since DB is equal to
the two sides AB, BD are equal to the two sides FB, BC respectively;
and the angle ABD is equal to the angle FBC;
therefore the base AD is equal to the base FC,
and the triangle ABD is equal to the triangle FBC. [I. 4]
Now the parallelogram BL
is double of the triangle ABD,
for they have the same base BD
and are in the same parallels
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.]
Therefore the parallelogram BL is also equal to the square GB.
the parallelogram CL can also be proved equal to the square HC;
therefore the whole square BDEC is equal to the two squares GB, HC. [C.N. 2]
And the square BDEC is described on BC,
and the squares GB, HC on BA, AC.
Therefore the square on the side BC is equal to the squares on the sides BA, AC.
Therefore etc. Q.E.D.
Book I: Euclid, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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