Euclid’s Theorem and Proof

0
Advertisement

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

Euclid's Theorem and Proof

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.

Advertisement
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}

Leave A Reply