Pythagoras's Theorem/Proof 7

Proof

 * [[File:Phytagoras.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:


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

where $\left({XYZ}\right)$ represents the area of $\triangle XYZ$.

This gives us:
 * $\dfrac {\left({ADC}\right)} {\left({ACB}\right)} = \dfrac {AC^2} {AB^2}$

and
 * $\dfrac {\left({CDB}\right)} {\left({ACB}\right)} = \dfrac {BC^2} {AB^2}$

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

Thus:
 * $\dfrac{\overbrace{\left({ADC}\right) + \left({CDB}\right)}^{\left({ACB}\right)}} {\left({ACB}\right)} = \dfrac {BC^2 + AC^2} {AB^2}$

This gives us $\therefore AB^2 = BC^2 + AC^2$ as desired.