Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 683.
Wednesday, November 2, 2011
Problem 683: Nagel Point, Triangle, Excircles, Tangency points, Concurrent lines, Semi-perimeter, Ceva theorem
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Right tr similar to each other
ReplyDeleteEcC'A to EbB'A, EbB'C to EaA'C, EaA'B to EcC'B
AB'/C'A=EbB'/EcC'
A'C/B'C=EaA'/EbB'
BC'/BA'=EcC'/EaA'
(AB'/C'A)(A'C/B'C)(BC'/BA')=(EbB'/EcC')(EaA'/EbB')(EcC'/EaA')
(BC'/C'A)(AB'/B'C)(CA'/BA')=1
AA', BB', CC' are concurent by ceva theorem