Euclid’s Theorem and Proof

Euclid's Theorem and Proof

We have similar triangles:

and we have also found two of Euclid’s theorems using similar triangles.

We call these statements Euclid’s right side theorems.

ΔDBA ∼ ΔDAC
$\displaystyle \Rightarrow \frac{\left| DA \right|}{\left| DC \right|}=\frac{\left| DB \right|}{\left| DA \right|}$
$\displaystyle \Rightarrow \frac{h}{k}=\frac{p}{h}\Rightarrow {{h}^{2}}=pk$

We call this statement Euclid’s altitude theorem.

We also have $\displaystyle \frac{\left| DB \right|}{\left| DA \right|}=\frac{\left| BA \right|}{\left| AC \right|}$
$\displaystyle \Rightarrow \frac{p}{h}=\frac{c}{b}\Rightarrow p=\frac{hc}{b}$
Now $\displaystyle {{c}^{2}}=ap\Rightarrow p=\frac{{{c}^{2}}}{a}$
so $\displaystyle \frac{{{c}^{2}}}{a}=\frac{hc}{b}\Rightarrow bc=ah$
A more simple way of finding this last formula is to look at the area of ΔABC in two different ways.
If we take [BC] as the base then [AD] is the altitude.

So the area of ΔABC is :

$\displaystyle A\left( ABC \right)=\frac{1}{2}x\left| BC \right|x\left| AD \right|=\frac{1}{2}ah$
If we take [AC] as the base then [AB] is the altitude. So the area of ΔABC is :
$\displaystyle A\left( ABC \right)=\frac{1}{2}x\left| AC \right|x\left| AB \right|=\frac{1}{2}bc$
$\displaystyle \Rightarrow \frac{1}{2}bc=\frac{1}{2}ah$
$\displaystyle \Rightarrow bc=ah$
Here are the formulas for Euclid’s theorems put together:
$\displaystyle {{c}^{2}}=ap,{{b}^{2}}=ak,{{h}^{2}}=ph,bc=ah$
Notice that if we write these formulas in the forms :
$\displaystyle \frac{a}{c}=\frac{c}{p},\frac{a}{b}=\frac{b}{k},\frac{p}{h}=\frac{h}{k},\frac{a}{b}=\frac{c}{h}$