Question

Consider the diagram and the paragraph proof below.

Answer 1

Given: Right △ABC as shown where CD is an altitude of the triangle. We  prove that $a^2 + b^2 = c^2$

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 $\frac{c}{a}$ and $\frac{a}{f}$ are true because they are ratios of corresponding parts of similar triangles.

The two proportions can be rewritten as $a^2 = cf$ and $b^2 = ce$.

Adding $b^2$ to both sides of first equation, $a^2 = cf$, results in the equation $a^2 + b^2 = cf + b^2$.

Because $b^2$ and ce are equal, ce can be substituted into the right side of the equation for $b^2$, resulting in the equation $a^2 + b^2 = cf + ce$.

Applying the converse of the distributive property results in the equation $a^2 + b^2 = c(f + e)$.

The last sentence of the proof is

Because f + e = c, $a^2 + b^2 = c^2$.