proposition 48
if the square on one of the side of a triangle is equal
to the (sum of the) squares on the two remaining side of
the triangle then the angle contained by the two remaining
sides of the triangle is a right-angle.
for let the square on one of the sides,BC , of triangle
ABC be equal to the (sum of the) squares on the sides
BA and AC . i say that angle BAC is a right-angle.
foe let AD have been drawn from point A at right-angles
to the triangle-line AC [prop.1.11] , and let AD
have been made equal to BA [prop.1.3], and let DC
have been joined. since DA is equal to AB, let the
square on AC have been added to both . thus, the (sum of the)
squares on BA and AC.but, the (square) on DC is
equal to the (sum of the square) on DA and AC . foe angle
DAC is a right-angle [prop.1.47]. but , the (square)
on BC is equal to (sum of the square) on BA and AC.
foe (that) was assumed.thus, the square on DC is equal
to the square on BC. sp side DC is also equal to (side)
BC. and since DA is equal AB, and AC (is) common
,the two (straight-lines) DA,AC are equal to the two
(straight-lines) BA, AC. and the base DC is equal
to the base BC . thus , angle DAC [is] equal to angle
BAC [prop.1.8] . but DAC is a right-angle. thus, BAC
is also a right-angle.
thus, if the square on one of the sides of a straight is
equal to the (sum of the) squares on the remaining two
sides of the triangle then the angle contained by the remaining
two sides of the triangle is a right-angle.,(whichis )
the very thing it was required to show.