Cross-Segment Identity
Revision as of 01:34, 2 January 2024 by Mathboy282 (talk | contribs) (Created page with "==Motivation== After repetitive usage of this in AMC/AIME Euclidean Geometry problems, I (mathboy282) only found it fit to make it an identity. ==Identity== Let there be a c...")
Motivation
After repetitive usage of this in AMC/AIME Euclidean Geometry problems, I (mathboy282) only found it fit to make it an identity.
Identity
Let there be a cyclic quadrilateral 
with 
as the intersection of the diagonals 
and 
Let 
and 
Then, we must have:
![\[ab+bc+cd+da=AB \cdot CD + BC \cdot DA.\]](http://latex.artofproblemsolving.com/3/b/8/3b8d4f729a446a1648f35d4287e6f782cae13bbb.png)
Proof
This comes as a direct result of Ptolemy's theorem.
![\[ab+bc+cd+da=b(a+c)+d(c+a)=(b+d)(a+c)=AB \cdot CD + BC \cdot DA.\]](http://latex.artofproblemsolving.com/f/a/9/fa904853c174b1833d681ccfcfe6b1e293a55394.png)
