MathItems

Proof P40 of T38

Right-angled triangle with hypotenuse c, legs a and b, and height h

Since the triangle CBP is similar to the triangle ABC, we have

ac=caa2=cc  ,\frac{a}{c} = \frac{c'}{a} \quad \Leftrightarrow \quad a^2 = c c' \; ,

and since the triangle ACP is similar to the triangle ABC, we have

bc=cbb2=cc  .\frac{b}{c} = \frac{c''}{b} \quad \Leftrightarrow \quad b^2 = c c'' \; .

We now have

a2+b2=cc+cc=c(c+c)=c2  .a^2 + b^2 = c c' + c c'' = c (c' + c'') = c^2 \; .