The exterior angle of a triangle equals the two opposite angles
Proof by parallels, citing the twenty-ninth proposition of Euclid
A page of Euclidean plane geometry built around a triangle lettered a c b. In a second view the base and one side are extended so that c becomes the exterior angle, and Leonardo states that exterior angle c is equivalent to the two opposite angles a and b. A further figure draws a triangle between two parallels, with equal angles marked by dots and dashes and justified 'by the first' and 'by the second of the twenty-ninth,' Euclid's proposition on parallels cut by a transversal. A closing note gives a rule for how much a polygon's angles surpass the first.
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Exterior angle equals the two opposite interior angles
With the base and one side of triangle a c b extended, c marks the exterior angle. Leonardo asserts that exterior angle c is equivalent to the two opposite angles, a and b.
Proof by the twenty-ninth proposition
A triangle set between two parallels on an extended base has equal angles marked by dots and dashes at f, c, a and b. The equalities are justified by the first and by the second of the twenty-ninth, Euclid's proposition on parallel lines cut by a transversal.
How a polygon's angles surpass the first figure
A concluding rule: take a figure of a given number of angles or sides, subtract two, and the remainder is the degree by which it surpasses the first. This measures how many triangles a polygon exceeds the base triangle by.