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 AD 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.
Book I: Euclid, Elements, Book I (ed. Sir Thomas L. Heath 1st Edition, 1908)
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