Ptolemy's theorem

 A quadrilateral is inscribed if and only if xy = ac + bd 

There is also the second Ptolemy relation :

 x/y = (ad + bc)/(ab + cd) 

Let's proof the direct theorem, that is when ABCD inscribed, then...
Recall the relations giving area of a triangle S = 1/2 b.c.sinA and S = a.b.c/(4R)
Area of quadrilateral is S = 1/2 x.y.sinθ (easy from area of the four internal triangles).

Draw CE // BD, hence triangles BCD and BED equal results into BE = c and DE = b
The corresponding angles AIB and ACE are equal, inscribed angles ABE and ACE are equal, hence ABE = θ
Area of ABCD = 1/2 x.y.sinθ is then the sum of areas of ABE and ADE, that is 1/2 a.c.sinθ + 1/2 b.d.sin(π - θ) hence
 xy = ac + bd 

Area(ABC) = a.b.x/(4R), Area(ACD) = c.d.x/(4R), Area(ABD) = a.d.y/(4R) and Area(BCD) = b.c.y/(4R) that is (ab + cd)x = (ad + bc)y and the second Ptolemy relation.


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