Додаток: Теорема Птолемея

Ptolemy's Theorem

In this post, we will prove the famous Ptolemy's theorem using the law of cosines.

Let us restate the theorem here.

Theorem (Ptolemy's theorem). In any quadrilateral $ABCD$, $$AB \cdot CD + AD \cdot BC = AC \cdot BD.$$

First, we must derive an expression for the lengths of the diagonals of the cyclic quadrilateral. Suppose that $a = AB$, $b = BC$, $c = CD$, $d=AD$, $e = AC$, and $f = BD$. In order to do this, apply the law of cosines on $\angle B$ and $\angle D$ and solve for $\cos \angle B$.

Let's check your result! Did you get $$e^2=a^2+b^2-2ab\cos B=c^2+d^2+2cd\cos B\\ \implies\cos B=\frac{a^2 + b^2 - c^2 - d^2}{2ab + 2cd}$$

Now plug this back in to the original equation for $e^2$.

You should've gotten: $$e^2 = \frac{(ac + bd)(ad+ bc)}{ab + cd}.$$

We plug this into our original equation to get $$\begin{align*} e^2 & = a^2 + b^2 - 2ab\left(\frac{a^2 + b^2 - c^2 - d^2}{2ab + 2cd}\right) \\ & = \frac{(a^2 + b^2)(ab + cd)}{ab + cd} - \frac{a^3b + b^3a - abc^2 - abd^2}{ab + cd} \\ & = \frac{(cda^2 + cdb^2 + abc^2 + abd^2}{ab + cd} \end{align*}$$ Now, we can see that $cda^2 + cdb^2 + abc^2 + abd^2 = ac(ad + bc) + bd(ad + bc) \\ = (ac + bd)(ad+ bc)$.

Thus $$e^2 = \frac{(ac + bd)(ad+ bc)}{ab + cd}.$$

What other expression must we have by symmetry?

We should get the following: $$f^2 = \frac{(ac + bd)(ab + cd)}{ad + bc}$$

The proof from here is now very simple. Note that $e^2 \cdot f^2 = (ac+ bd)^2.$ Thus, $ef = ac + bd$. $\Box$

Ptolemy's Theorem

Suppose we had a cyclic quadrilateral $ABCD$. As cyclic quadrilaterals are special, we want to find a nice relationship between the sides of the quadrilateral.

How do we do this? Well, since most of what we know about cyclic quadrilaterals is in angles, we try to find similar triangles:

We can immediately see that $\triangle ABP \sim \triangle DCP$. However, trying to use similarity ratios on this won't tell us much more that what we already know. Instead, we draw point $K$ on $AC$ such that $\angle ABK \cong \angle DBC$.

By similar triangles (AA), we have: $$\triangle ABK \sim \triangle DBC$$ $$\triangle ABD \sim \triangle KBC$$ From similarity, we have: $$AK \cdot BD = AB \cdot CD$$ $$CK \cdot BD = AD \cdot BC$$ Adding these up, we get: $$BD(AK + CK) = AB \cdot CD + AD \cdot BC$$ $$BD \cdot AC = AB \cdot CD + AD \cdot BC$$ And that's it. We've found a relation! What we have just proven is known as Ptolemy's Theorem. It can also be stated as:

The sum of the product of opposite sides of a cyclic quadrilateral is equal to the product of the diagonals.

Note: there is a variant of Ptolemy's Theorem for all cyclic quadrilaterals called Ptolemy's Inequality. It states $AB\cdot CD + AD \cdot BC \ge AC \cdot BD$.

That's all there is to Ptolemy's Theorem. Thanks for reading!