Pythagoras's Theorem/Proof 7

Theorem

Let $\triangle ABC$ be a right triangle with $c$ as the hypotenuse.

Then:

$a^2 + b^2 = c^2$

Proof

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 by Ratio of Areas of Similar Triangles:

$\dfrac {\paren {XYZ} } {\paren {X' Y' Z'} } = \dfrac {XY^2} {X'Y'^2}$

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.

$\blacksquare$

Source of Name

This entry was named for Pythagoras of Samos.