Is This Proof of Pythagorean Theorem?

  1. Given image:
  2. $trianglemathit{BDE}$ is an altitude right triangle as given by the image.
  3. $trianglemathit{BDE}$ and $trianglemathit{FDE}$ both contain 90° given by the image; therefore, $trianglemathit{BDE}$ and $triangle{mathit{FDE}}$ are both right triangles.
  4. $trianglemathit{BDE}$ is an altitude right triangle with $trianglemathit{FDE}$ as given by the image.
  5. If $trianglemathit{BDE}$ is an altitude right triangle with $trianglemathit{FDE}$ then:
    • $overline{mathit{DE}}^{2}=overline{mathit{BE}}overline{mathit{EF}}$
    • $overline{mathit{DF}}^{2}=overline{mathit{EF}}overline{mathit{BF}}$
  6. $overline{mathit{BE}}+overline{mathit{EF}}=overline{mathit{BF}}$ as given by the image
  7. Given these equations algebra can then be performed to solve for $overline{mathit{DF}}$. Algebra:
    • Rearranging:$overline{mathit{DE}}^{2}=overline{mathit{BE}}overline{mathit{EF}}$ to get $frac{overline{mathit{DE}}^{2}}{overline{mathit{EF}}}=overline{mathit{BF}}.$
    • $overline{mathit{DF}}^{2}=overline{mathit{EF}}overline{mathit{BF}}$
    • Substituting:$$overline{mathit{DF}}^{2}=overline{mathit{EF}}(overline{mathit{BE}}+overline{mathit{EF}})$$
    • Substituting Again:$$overline{mathit{DF}}^{2}=overline{mathit{EF}}(frac{overline{mathit{DE}}^{2}}{overline{mathit{EF}}} + overline{mathit{EF}})$$
    • Simplifying:$$overline{mathit{DF}}^{2}=overline{mathit{DE}}^{2}+overline{mathit{EF}}^{2}$$
  • Which is the familiar form of Pythagorean Theorem.
  • Is this a mathematically correct derivation?
    Is this proof general enough to be used as a general proof?