Pythagoras's Theorem/Proof 7

Proof

 * Pythagoras7.png

Let $\triangle ABC$ be a right triangle and $h_c$ the altitude from $c$.

We have:


 * $\angle CAB \cong \angle DCB$


 * $\angle ABC \cong \angle ACD$

Then we have:


 * $\triangle ADC \sim \triangle ACB \sim \triangle CDB$

Use the fact that if $\triangle XYZ \sim \triangle X'Y'Z'$ then:


 * $\dfrac {\paren {XYZ} } {\paren {X' Y' Z'} } = \dfrac {XY^2} {X'Y'^2} = \dfrac {h_z^2} {h_{z'}^2} = \dfrac {t_z^2} {t_{z'}^2} = \ldots$

where $\paren {XYZ}$ denotes the area of $\triangle XYZ$.

This gives us:
 * $\dfrac {\paren {ADC} } {\paren {ACB} } = \dfrac {AC^2} {AB^2}$

and
 * $\dfrac {\paren {CDB} } {\paren {ACB} } = \dfrac {BC^2} {AB^2}$

Taking the sum of these two equalities we obtain:
 * $\dfrac {\paren {ADC} } {\paren {ACB} } + \dfrac {\paren {CDB} } {\paren {ACB} } = \dfrac {BC^2} {AB^2} + \dfrac {AC^2} {AB^2}$

Thus:
 * $\dfrac {\overbrace {\paren {ADC} + \paren {CDB} }^{\paren {ACB} } } {\paren {ACB} } = \dfrac {BC^2 + AC^2} {AB^2}$

Thus:
 * $AB^2 = BC^2 + AC^2$

as desired.