# Ptolemy’s Theorems

**About
**

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

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,

since their sum is

We will use that and the formulas

,

(*Product to Sum formula),*

*(Sum to Product formula)*

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

(1)

Similarly, (2)

So

But

(3)

And

Furthermore

(4)

And in the same way (5)

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

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:

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 and (ABCD).

Take E, concyclic with points A,B,C,D such that . So implying that where is the arc “x”.

It is and using this,

(since )

.

So we conclude that

**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 (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 , , . Introdusing these in (1) we get

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

**Proof**

In every triangle with sides and the radius of the circumcircle of the triangle and the height to side a respectively, it is known that . So, according to the figure,

(1)

And similarly (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,

with equality if and only if ABCD is cyclic.

- Proof of direct Pythagorean Theorem. Let ABDC be a rectangle, obviously a cyclic quadrilateral. By Ptolemy’s we obtain

- 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 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 and

**Exercises**

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