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 $$\frac{(XYZ)}{(X'Y'Z')}=\frac{XY^2}{X'Y'^2}=\frac{h_z^2}{h_{z'}^2}=\frac{t_z^2}{t_{z'}^2}=...$$ where $$(XYZ)$$ represents the area of $$\triangle XYZ$$.

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

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

Thus, $$\frac{\overbrace{(ADC)+(CDB)}^{(ACB)}}{(ACB)}=\frac{BC^2+AC^2}{AB^2}$$.

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.
 * $$\frac {AC}{AB} = \frac {AD}{AC}$$
 * $$\frac {BC}{AB} = \frac {BD}{BC}$$

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.