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;
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
BC, and
FB to
BA, 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
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.]
Therefore the parallelogram BL is also equal to the square GB.
Similarly, if
AE,
BK be joined, 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.
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