# 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?