Equal squares on an isosceles right triangle: a congruence proof
A full demonstration that the squares on the sides are equal, citing 'the fourth of the first'
The folio gives a full geometric demonstration for an isosceles right triangle whose squares on the sides are divided by their diagonals. Leonardo argues that two triangles are equal because their bases (e a and a b) are sides of the lower square a b o e and their sides (d a and a c) are sides of the larger square d a c h, with the angles e a b and d a c both right; adding triangle e a n to each and invoking 'the fourth of the first' (Euclid I.4) makes side d b equal to side e c. He then proves the squares n a m c and a b e o equal, each being double its triangle 'by the twenty-third,' whose bases are common to triangle and square. Overlapping-triangle figures and a square-and-parallelogram diagram at the right illustrate the steps.
On this page
Two partially overlapping triangles (b e, a d, c)
The construction sets up two triangles that partly overlap, labelled with b e, a d and c. These are the figures whose equality the demonstration establishes.
Parallelogram and square applied to the triangle
A parallelogram and a square are applied to the triangle, carrying the labels o, b e, a n d and c m h. This figure ties the square on the side to the triangle used in the proof.
The two triangles are equal (bases and sides from the squares)
Base e a equals base a b as sides of the square a b o e, and side d a equals side a c as sides of the larger square d a c h, with angles e a b and d a c both right. Adding triangle e a n to each and citing 'the fourth of the first,' side d b is proved equal to side e c.
The squares n a m c and a b e o are equal ('the twenty-third')
Each square is shown to be double its triangle 'by the twenty-third,' which makes their bases common to both triangle and square. From this the two squares n a m c and a b e o are proved equal.