Proposed Problem

Click the figure below to see the complete problem 426 about Triangle, Circumradius, Circumcenter, concurrent Cevians.

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Complete Problem 426

Level: High School, SAT Prep, College geometry

## Monday, February 15, 2010

### Problem 426: Triangle, Circumradius, Circumcenter, concurrent Cevians

Labels:
cevian,
circumcenter,
circumcircle,
circumradius,
concurrent,
triangle

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from ceva's theorem

ReplyDeleteOD/AD + OE/BE + OF/FC = 1

(AD-R)/AD + (BE-R)/BE + (FC-R)Fc = 1

1 - R/AD + 1 - R/BE + 1 - R/FC = 1

2 = R/AD + R/BE + R/FC

2 = R ( 1/AD + 1/BE + 1/FC )

2/R = 1/AD + 1/BE + 1/FC

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This is a special case of Ceva:

ReplyDeleteFor any point P in ABC, if we draw three cevians APD, BPE, and CPF, then

AP/AD + BP/BE + CP/CF = 2