proof of Ptolemy’s theorem
Let ABCD be a cyclic quadrialteral. We will prove that
AC⋅BD=AB⋅CD+BC⋅DA. |
Find a point E on BD such that ∠BCA=∠ECD. Since ∠BAC=∠BDC for opening the same arc, we have triangle similarity
△ABC∼△DEC and so
ABDE=CACD |
which implies AC⋅ED=AB⋅CD.
Also notice that △ADC∼△BEC since have two pairs of equal angles. The similarity implies
ACBC=ADBE |
which implies AC⋅BE=BC⋅DA.
So we finally have AC⋅BD=AC(BE+ED)=AB⋅CD+BC⋅DA.
Title | proof of Ptolemy’s theorem |
---|---|
Canonical name | ProofOfPtolemysTheorem |
Date of creation | 2013-03-22 12:38:31 |
Last modified on | 2013-03-22 12:38:31 |
Owner | drini (3) |
Last modified by | drini (3) |
Numerical id | 11 |
Author | drini (3) |
Entry type | Proof |
Classification | msc 51-00 |
Related topic | PtolemysTheorem |
Related topic | CyclicQuadrilateral |