# Proof of Pythagoras' Theorem

The following is due to Dao Thanh Oai (Vietnam):

Let $CH_c$ and $BH_b$ be two altitudes in $\Delta ABC.$

Then

$BC^{2}=AB\cdot BH_{c}+AC\cdot CH_{b}.$

### Proof

Denoting angles at the vertices of the triangle $\alpha ,\beta ,\gamma$ we find several lengths as shown in the diagram below:

We, therefore, have:

\begin{align} AB\cdot BH_{c}+AC\cdot CH_{b} &= AB\cdot BC\cdot\cos\beta + AC\cdot BC\cdot\cos\gamma\\ &=BC(AB\cdot\cos\beta+AC\cdot\cos\gamma)\\ &=BC\cdot(BH_a + CH_a)\\ &=BC\cdot BC = BC^{2}. \end{align}

We could have done that without trigonometry but from similutude of triangles:

$\displaystyle\frac{BH_c}{BC}=\frac{BH_a}{AB}$ and $\displaystyle\frac{CH_b}{BC}=\frac{CH_a}{AC}$

It follows that

\begin{align} AB\cdot BH_{c}+AC\cdot CH_{b} &= BC\cdot BH_a + BC\cdot CH_a\\ &=BC\cdot(BH_a + CH_a)\\ &=BC\cdot BC = BC^{2}. \end{align}

The Pythagorean theorem is obtained when angle at $A$ is right.