## Pythagoras theorem, demonstration – one of many – extra

Posted in Math Theory by Lia on 06/01/2017

### Proof that:

$\Delta&space;\textbf{A}_{\textbf{1}}\textbf{A}_{\textbf{2}}\textbf{D}_{\textbf{2}}\textbf{=}\Delta&space;\textbf{B}_{\textbf{1}}\textbf{B}_{\textbf{2}}\textbf{A}_{\textbf{2}}\textbf{=}\Delta&space;\textbf{C}_{\textbf{1}}\textbf{C}_{\textbf{2}}\textbf{B}_{\textbf{2}}\textbf{=}\Delta&space;\textbf{D}_{\textbf{1}}\textbf{D}_{\textbf{2}}\textbf{C}_{\textbf{2}}$

### 1. all four triangles have one of the side equal:

$\textbf{A}_{\textbf{2}}\textbf{B}_{\textup{2}}\textbf{=}\textbf{B}_{\textbf{2}}\textbf{C}_{\textup{2}}\textbf{=}\textbf{C}_{\textbf{2}}\textbf{D}_{\textup{2}}\textbf{=}\textbf{D}_{\textbf{2}}\textbf{A}_{\textup{2}}\textbf{=}\textbf{a}$

### This means that:

$\Delta&space;\textbf{A}_{\textbf{1}}\textbf{A}_{\textbf{2}}\textbf{D}_{\textbf{2}}\textbf{=}\Delta&space;\textbf{B}_{\textbf{1}}\textbf{B}_{\textbf{2}}\textbf{A}_{\textbf{2}}\textbf{=}\Delta&space;\textbf{C}_{\textbf{1}}\textbf{C}_{\textbf{2}}\textbf{B}_{\textbf{2}}\textbf{=}\Delta&space;\textbf{D}_{\textbf{1}}\textbf{D}_{\textbf{2}}\textbf{C}_{\textbf{2}}$

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