Geometry Problem

Click the figure below to see the complete problem 488 about Triangle, Cevian, Concurrency, Circle, Circumcircle.

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Complete Problem 488

Level: High School, SAT Prep, College geometry

## Saturday, July 31, 2010

### Problem 488: Triangle, Cevian, Concurrency, Circle, Circumcircle

Labels:
cevian,
circle,
circumcircle,
concurrent,
triangle

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Let n=BA’ .BA” = BC’.BC” ( circumcircle A’B’C’ and secants BC’C” , BA”A’ )

ReplyDeleteand m=AB’.AB”=AC’.AC” ; p=CA’.CA”=CB”.CB’ ( same reasons as above)

1. Apply Ceva’s theorem for triangle ABC and point of concurrent D

(A’B/A’C) * (B’C/B’A) *(C’A/C’B)=1

2. Replace A’B=m/A”B ; A’C=p/A”C etc…. in above equation we have :

(m/A”B)/(p/A”C) * (p/B”C)/(m/B”A) * ( m/C”A)/(n/C”B) =1

3. Simplify above equation we get (A”C/A”B )*(B”A/B”C)*(C”B/C”A) =1

Per Ceva’s theorem AA”, BB”, CC” are concurrent.

Peter Tran

vstran@yahoo.com

The concurrency point is called "the cyclocevian conjugate of P".

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