Pythagoras's Theorem/Proof 2

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

where $(XYZ)$ represents the area of $\triangle XYZ$.

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

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

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

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

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

Alternative Derivation
Again from the above diagram, we have:
 * $\dfrac {AC}{AB} = \dfrac {AD}{AC}$
 * $\dfrac {BC}{AB} = \dfrac {BD}{BC}$

using the fact that all the triangles involved are similar.

That is:
 * $AC^2 = AB.AD$
 * $BC^2 = AB.BD$

Adding, we now get: $AC^2 + BC^2 = AB.AD + AB.BD = AB (AD + BD) = AB^2$

Historical Note
This proof was demonstrated by in the 12th century.

It was rediscovered in the 17th century by.