Pythagoras's Theorem/Proof 2

Theorem
Given any right triangle $\triangle ABC$ with $c$ as the hypotenuse, we have $a^2 + b^2 = c^2$.

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

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

This gives us:
 * $\displaystyle \frac {(ADC)} {(ACB)} = \frac {AC^2} {AB^2}$ and $\displaystyle \frac {(CDB)} {(ACB)} = \frac {BC^2} {AB^2}$

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

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

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

Alternative Derivation
Again from the above diagram, we have: using the fact that all the triangles involved are similar.
 * $\displaystyle \frac {AC}{AB} = \frac {AD}{AC}$
 * $\displaystyle \frac {BC}{AB} = \frac {BD}{BC}$

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 Bhāskara II Āchārya in the 12th century.

It was rediscovered in the 17th century by John Wallis.