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Ptolemy’s Theorems

August 6, 2010

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

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)
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One Comment
  1. Hi, I log on to your new stuff daily. Your story-telling style is awesome, keep doing what you’re doing!

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