Proposition 48.
If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.
For in the triangle
ABC let the square on one side
BC be equal to the squares on the sides
BA,
AC;
I say that the angle
BAC is right.
For let
AD be drawn from the point
A at right angles to the straight line
AC, let
AD be made equal to
BA, and let
DC be joined.
Since
DA is equal to
AB, the square on
DA is also equal to the square on
AB.
Let the square on
AC be added to each;
therefore the squares on DA, AC are equal to the squares on BA, AC.
But the square on
DC is equal to the squares on
DA,
AC, for the angle
DAC is right; [
I. 47] and the square on
BC is equal to the squares on
BA,
AC, for this is the hypothesis;
therefore the square on DC is equal to the square on BC, so that the side DC is also equal to BC.
And, since
DA is equal to
AB, and
AC is common,
the two sides DA, AC are equal to the two sides BA, AC; and the base
DC is equal to the base
BC;
therefore the angle DAC is equal to the angle BAC. [I. 8] But the angle
DAC is right;
therefore the angle BAC is also right.
Therefore etc.
Q. E. D.