Consider the diagram and the paragraph proof below. Given: Right △ABC as shown where CD is an altitude of the triangle
Prove: a2 + b2 = c2

Triangle A B C is shown. Angle A C B is a right angle. An altitude is drawn from point C to point D on side A B to form a right angle. The length of C B is a, the length of A C is b, the length of A B is c, the length of A D is e, the length of D B is f, and the length of C D is h.

Because △ABC and △CBD both have a right angle, and the same angle B is in both triangles, the triangles must be similar by AA. Likewise, △ABC and △ACD both have a right angle, and the same angle A is in both triangles, so they also must be similar by AA. The proportions StartFraction c Over a EndFraction = StartFraction a Over f EndFraction and StartFraction c Over b EndFraction = StartFraction b Over e EndFraction are true because they are ratios of corresponding parts of similar triangles. The two proportions can be rewritten as a2 = cf and b2 = ce. Adding b2 to both sides of first equation, a2 = cf, results in the equation a2 + b2 = cf + b2. Because b2 and ce are equal, ce can be substituted into the right side of the equation for b2, resulting in the equation a2 + b2 = cf + ce. Applying the converse of the distributive property results in the equation a2 + b2 = c(f + e).

Which is the last sentence of the proof?

Because f + e = 1, a2 + b2 = c2.
Because f + e = c, a2 + b2 = c2.
Because a2 + b2 = c2, f + e = c.
Because a2 + b2 = c2, f + e = 1.