#### What is Euclid’s Theorem for triangles? The proof of Euclid’s theorem and examples. ### Euclid’s Theorem;

We have similar triangles:

• ΔABC ∼ ΔDBA ∼ ΔDAC

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

• c²=ap and b²=ak

We call these statements Euclid’s right side theorems.

###### Here are Euclid’s other theorems on right angled triangles :
• 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 :
• $\displaystyle \frac{a}{c}=\frac{c}{p},\frac{a}{b}=\frac{b}{k},\frac{p}{h}=\frac{h}{k},\frac{a}{b}=\frac{c}{h}$

then :

• c is the mean proportional between a and p
• b is the mean proportional between a and k
• h is the mean proportional between p and k
• h is the fourth proportional to a, b and c

### SUMMARY

• $\displaystyle {{c}^{2}}=ap\Rightarrow c=\sqrt{ap}$
• $\displaystyle {{h}^{2}}=pk\Rightarrow h=\sqrt{pk}$
• $\displaystyle {{b}^{2}}=ak\Rightarrow b=\sqrt{ak}$
• $\displaystyle ah=bc\Rightarrow h=\frac{bc}{a}$