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

Labels:
center,
Ceva's theorem,
concurrent,
excircle,
Nagel theorem,
semiperimeter,
tangency point,
triangle

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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