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.
Proof
Erect squares on each side via I.46; using I.14, I.31 and I.41 show
each part-square equals a corresponding parallelogram cut off the
hypotenuse-square by the perpendicular from the right angle; sum via
Common Notion 2.
Figure
Proposition I.47. The square on the hypotenuse BC is partitioned by the altitude AH extended into two rectangles, each equal (by I.41 + I.46) to a square on a leg; thus BC2=AB2+AC2.