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Kosnita's Theorem: Circumcenters

Level: High School, SAT Prep, College geometry

## Thursday, November 26, 2009

### Kosnita's Theorem: Circumcenters, Concurrent lines

Labels:
center,
circumcenter,
circumcircle,
concurrent,
Kosnita,
triangle

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∠AOB=2∠ACB=2∠C

ReplyDelete∠AOCB=2(180°-∠AOB)=360°-4∠C

∠OCAB=(180°-∠AOCB)/2=2∠C-90°

Let C' be the point of intersection of AB and COC.

Area of ΔCAOC=1/2×AC×AOC×sin(A+∠OCAB)=1/2×AC×AOC×sin(A+2C-90°)=1/2×AC×AOC×sin(90°+C-B)=1/2×AC×AOC×cos(C-B)

AC'/C'B=Area of ΔCAOC/Area of ΔCBOC=1/2×AC×AOC×cos(C-B)/(1/2×BC×AOC×cos(C-A))=AC×cos(C-B)/(BC×cos(C-A))

AC'/C'B×BA'/A'C×CB'/B'A=1