PlanetMath: proof of Ptolemy's theorem

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (232)
Orphanage
Unclass'd
Unproven (519)
Corrections (22)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

proof of Ptolemy's theorem

(Proof)

Let be a cyclic quadrialteral. We will prove that

$\begin{displaymath}AC\cdot BD=AB\cdot CD + BC\cdot DA.\end{displaymath}$

$\includegraphics{ptolomeo}$

Find a point on such that $\angle BCA=\angle ECD$ . Since $\angle BAC=\angle BDC$ for opening the same arc, we have triangle similarity $\triangle ABC\sim \triangle DEC$ and so

$\begin{displaymath}\frac{AB}{DE}=\frac{CA}{CD}\end{displaymath}$

which implies $AC\cdot ED = AB\cdot CD$ .

Also notice that $\triangle ADC \sim \triangle BEC$ since have two pairs of equal angles. The similarity implies

$\begin{displaymath}\frac{AC}{BC}=\frac{AD}{BE}\end{displaymath}$

which implies $AC\cdot BE = BC\cdot DA$ .

So we finally have $AC\cdot BD=AC(BE+ED)=AB\cdot CD+BC\cdot DA$ .

"proof of Ptolemy's theorem" is owned by drini. [ owner history (1) ]

(view preamble | get metadata)

See Also: Ptolemy's theorem, cyclic quadrilateral

This object's parent.

Log in to rate this entry.
(view current ratings)

Cross-references: angles, implies, similarity, triangle, arc, point, cyclic

This is version 8 of proof of Ptolemy's theorem, born on 2002-05-15, modified 2002-05-16.
Object id is 2906, canonical name is ProofOfPtolemysTheorem.
Accessed 8180 times total.

Classification:

AMS MSC:

51-00 (Geometry :: General reference works )

Pending Errata and Addenda

None.

Discussion

forum policy

No messages.

Interact

post | correct | update request | add example | add (any)