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[parent] proof of Ptolemy's inequality (Proof)

Looking at the quadrilateral $ ABCD$ we construct a point $ E$, such that the triangles $ ACD$ and $ AEB$ are similar ( $ \angle ABE=\angle CDA$ and $ \angle BAE=\angle CAD$).

\includegraphics{ptolomaeus.eps}
This means that:

$\displaystyle \frac{AE}{AC}=\frac{AB}{AD}=\frac{BE}{DC},$
from which follows that

$\displaystyle BE=\frac{AB\cdot DC}{AD}.$
Also because $ \angle EAC=\angle BAD$ and

$\displaystyle \frac{AD}{AC}=\frac{AB}{AE}$
the triangles $ EAC$ and $ BAD$ are similar. So we get:

$\displaystyle EC=\frac{AC\cdot DB}{AD}.$
Now if $ ABCD$ is cyclic we get

$\displaystyle \angle ABE+\angle CBA=\angle ADC+\angle CBA=180^\circ.$
This means that the points $ C$, $ B$ and $ E$ are on one line and thus:

$\displaystyle EC=EB+BC$
Now we can use the formulas we already found to get:

$\displaystyle \frac{AC\cdot DB}{AD}=\frac{AB\cdot DC}{AD}+BC.$
Multiplication with $ AD$ gives:

$\displaystyle AC\cdot DB=AB\cdot DC+BC\cdot AD.$

Now we look at the case that $ ABCD$ is not cyclic. Then

$\displaystyle \angle ABE+\angle CBA=\angle ADC+\angle CBA\neq 180^\circ,$
so the points $ E$, $ B$ and $ C$ form a triangle and from the triangle inequality we know:

$\displaystyle EC<EB+BC.$
Again we use our formulas to get:

$\displaystyle \frac{AC\cdot DB}{AD}<\frac{AB\cdot DC}{AD}+BC.$
From this we get:

$\displaystyle AC\cdot DB<AB\cdot DC+BC\cdot AD.$
Putting this together we get Ptolomy's inequality:

$\displaystyle AC\cdot DB\leq AB\cdot DC+BC\cdot AD,$
with equality iff $ ABCD$ is cyclic.



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Cross-references: iff, equality, inequality, triangle inequality, multiplication, formulas, line, cyclic, similar, triangles, point, quadrilateral

This is version 2 of proof of Ptolemy's inequality, born on 2002-06-09, modified 2006-10-02.
Object id is 3079, canonical name is ProofOfPtolemysInequality.
Accessed 8213 times total.

Classification:
AMS MSC: 51-00 (Geometry :: General reference works )
Pending Errata and Addenda
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