PROPOSITION 47. THEOREM.
In any right-angled triangle, the square which is de-scribed on the side subtending the right angle is equal to the squares described on the sides which contain the right angle.
Let ABC be a right-angled triangle, having the right angle BAC : the square described on the side BC shall be equal to the squares described on the sides BA, AC.
On BC describe
the square BDEC,
and on BA, AC de-
scribe the squares
GB,HC; [I.46.
through A draw AL
parallel to BD or
CE ; [I. 31.
and join AD, FC.
Then, because the angle BAC is a right angle, [Hypothesis.
and that the angle
BAG is also a right
angle, [Definition 30.
the two straight lines AC, AG, on the opposite sides of
AB, make with it at the point A the adjacent angles equal
to two right angles ;
therefore CA is in the same straight line with AG. [I, 14.
For the same reason, AB and AH are in the same straight line.
Now the angle DBC is equal to the angle FBA, for each
of them is a right angle. [Axiom 11.
Add to each the angle ABC.
Therefore the whole angle DBA is equal to the whole angle
FBC. [Axiom 2.
And because the two sides AB, BD are equal to the two
sides FB, BC, each to each ; [Definition 30.
and the angle DBA is equal to the angle FBC;
therefore the triangle ABD is equal to the triangle FBC. [I. 4.
