Found by Ptolemy of Alexandria (or Claudius Ptolemaeus) (90-168 AD).

Ptolemy’s First Theorem

A quadrilateral ABCD is cyclic (inscribed) if and only if

$AB\cdot CD + AD\cdot BC=AC\cdot BD$

or

“A quadrilateral is inscribed in a circle if and only if the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals”

Proofs

Proof 1 of the direct theorem(Using a little Trigonometry :))

As in the figure, $a=a_1, b=b_1, c=c_1, d=d_1,$

$\sin(a+b)= \sin(d+c_1),$

$\sin(c+a_1)= \sin(b_1+d_1)$ since their sum is $\pi$

We will use that $\cos X=- \cos(180^\circ-X)$ and the formulas

$\cos(90^\circ-X)= \sin X$,

$2 \sin X\cdot \sin Y= \cos(X-Y)- \cos(X+Y)$ (Product to Sum formula),

$\cos X+ \cos Y=2\cdot \cos \frac{X+Y}{2}\cdot \cos \frac{X-Y}{2}$ (Sum to Product formula)

Applying the extended law of sines to triangles  ADB, DBC respectively and the Sum to Product formula we obtain:

$AB\cdot CD=(2R \sin d) \cdot (2R\sin a)=2R^2( \cos(d-a) - \cos(d+a))$ (1)

Similarly, $AD\cdot BC=4R^2 \sin c\cdot \sin b=2R^2(\cos(c-b) - \cos(c+b))$ (2)

So $AB\cdot CD + AD\cdot BC=$

$=2R^2(\cos(d-a) - \cos(d+a) + \cos(c-b) - \cos(c+b) )$

But $a+b+c+d=180^\circ \Rightarrow d+a=180^\circ -(b+c) \Rightarrow$

$\Rightarrow \cos(d+a)= - \cos(c+b) \Rightarrow \cos(d+a) + \cos(b+c)=0$(3)

And $\cos(d-a) + \cos(c-b)=2\cdot \cos(\frac{d+c-a-b}{2})\cdot \cos(\frac{d-a-c+b}{2})$

Furthermore

$d+c=180^\circ-a-b \Rightarrow \frac{d+c-a-b}{2}=90^\circ-a-b \Rightarrow$

$\Rightarrow \cos(\frac{d+c-a-b}{2})= \sin(a+b)$ (4)

And in the same way $\cos(\frac{d-a-c+b}{2})= \sin(a+c)$ (5)

Combining the results 3, 4 and 5 and using again the Law of Sines:

$AB\cdot CD + AD\cdot BC=2R^2(2 \sin(a+b)\cdot \sin(a+c))=$

$=2R \sin(a+b)\cdot 2R \sin(a+c)=BC\cdot AD$

And the proof is completed.

Proof 2 of the direct theorem (With Areas)

We know that in a convex quadrilateral ABCD, its area can be given by:

$(ABCD)= \frac{1}{2}AC \cdot BD \cdot \sin(APB)$

where P is the intersection of AC and BD and from now on S(x) will present the area of polygon “x”.

We will find a relation between $AB\cdot CD + AD\cdot BC$ and (ABCD).

Take E, concyclic with points A,B,C,D such that $BE\parallel AC$. So $AB=EC, AE=BC$ implying that $\widetilde{AB} =\widetilde{EC}$ where $\widetilde{x}$ is the arc “x”.

It is $\widehat{EAD}= \frac{ \widetilde{EC}+ \widetilde{CD}}{2}= \frac{ \widetilde{AB}+ \widetilde{CD}}{2}= \widehat{APB}$ and using this,

$(ABCD)=(ABC)+(ACD)=$

$=(AEC)+(ACD)=$

$=(AECD)=(AED)+(ECD)$

$\Leftrightarrow (ABCD)= \frac{1}{2}AE \cdot AD \sin (EAD)+ \frac{1}{2}EC \cdot CD \sin ECD =$

$= \frac{1}{2} \sin{EAD} \cdot(AE \cdot AD + EC \cdot CD)$ (since $\widehat{EAD}+ \widehat{ECD}=180^\circ$)

$= \frac{1}{2} \sin{EAD} \cdot (BC \cdot AD + AB \cdot CD)$.

So we conclude that  $(ABCD)= \frac{1}{2}AC \cdot BD \cdot \sin APB= \frac{1}{2} \sin{EAD} \cdot(BC \cdot AD + AB \cdot CD)$

$\Leftrightarrow AB\cdot CD + AD\cdot BC=AC\cdot BD$

Proof of the converse of the theorem (By Inversion)

For the prove of the converse of Ptolemy’s 1st Theorem we will work with inversion. Suppose that $AB\cdot CD + AD\cdot BC=AC\cdot BD$ (1). We will prove that ABCD is inscribed.

Consider the inversion with center A and any radius r (suppose r<min{AD,AB} as in the figure). Let the images of B,C,D be B’,C’,D’ respectively. By inversion, triangles ABC-AC’B’, ACD-AD’C’ and ABD-AD’B’ are similar which implies that $BC= \frac{AB \cdot AC}{r^2} \cdot B'C'$, $CD= \frac{AC \cdot AD}{r^2} \cdot C'D'$, $BD= \frac{AB \cdot AD}{r^2} \cdot B'D'$. Introdusing these in (1) we get

$AD \cdot \frac{AB \cdot AC}{r^2}B'C'+AB \cdot \frac{AC \cdot AD}{r^2}C'D' = AC \cdot \frac{AB \cdot AD}{r^2}B'D'$ $\Leftrightarrow B'C'+C'D'=B'D$ implying that B’, C’ and D’ are collinear. By inversion, this means that points B, C and D are concyclic, so the proof is completed.

Ptolemy’s Second Theorem

If a quadrilateral ABCD is cyclic then, for the diagonals AC and BD, it is true that

$\frac{AC}{BD}= \frac{AB\cdot AD + CB\cdot CD}{BA\cdot BC + DA\cdot DC}$

Proof

In every triangle with sides $a, b, c$ and $R, h_a$ the radius of the circumcircle of the triangle and the height to side a respectively, it is known that $bc=h_a \cdot 2R$. So, according to the figure,

$ad+bc=2Rh_1+2Rh_2=$

$=2R \sin w \cdot AK+2R \sin w \cdot KC$

$=2R \cdot \sin w \cdot AC$ (1)

And similarly $ab+cd=2R \cdot \sin( 180^\circ -w) \cdot BD$ (2). Dividing 1 with 2 we get the desired relation.

Notes

– By Ptolemy’s theorem, we can produce Ptolemy’s Inequality:

For every quadrilateral ABCD, $AB\cdot CD + AD\cdot BC \ge AC\cdot BD$

with equality if and only if ABCD is cyclic.

Results

• Proof of direct Pythagorean Theorem. Let ABDC be a rectangle, obviously a cyclic quadrilateral. By Ptolemy’s we obtain $AB \cdot CD + BD \cdot AC = AD \cdot BC$ $\Leftrightarrow AB^2 + AC^2 = BC^2$
• Let ABC be an equilateral triangle of side a and P a point in its plane. Then P belongs to the circumcircle of the triangle if and only if PB=PA+PC. We have equivalently           $PB=PA+PC \Leftrightarrow aPB=aPA+aPC \Leftrightarrow$ ABCP is inscribed.
• Mahavira’s Theorem In a cyclic quadrilaral with sides a, b, c, d  and diagonals m, n (as in the figure) the length of the diagonals are given by $m^2= \frac{(ab+cd)(ac+bd)}{ad+bc}$ and $n^2= \frac{(ad+bc)(ac+bd)}{(ab+cd)}$

Exercises

1. If a point P lies on the arc CD of the circumcircle of a square ABCD, then PA(PA+PC)=PB(PB+PD)