See complete Problem 130

Triangle, Concurrent Cevians, Sum of Ratios. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Wednesday, June 25, 2008

### Elearn Geometry Problem 130

Labels:
cevian,
concurrent,
ratio,
triangle

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can I know what is a cevian, please

ReplyDeleteA Cevian is any line segment which joins a vertex of a triangle with a point on the opposite side (or its extension).

ReplyDeletefrom an cevian theorem

ReplyDeletePD/AD + PE/BE + PF/FC = 1

=>

(AD - PA)/AD + (BE - PB)/BE + (FC - PC)/FC = 1

AD/AD - PA/AD + BE/BE - PB/BE + FC/FC - PC/FC = 1

1 - PA/AD + 1 - PB/BE + 1 - PC/FC = 1

PA/AD + PB/BE + PC/FC = 2

-------------------------------------------

PA/AD=PBA/ABD=PCA/ACD=(PBA+PCA)/(ABD+ACD)=(PBA+PCA)/ABC

ReplyDeletePB/BE=PBC/BEC=PBA/BEA=(PBC+PBA)/(BEC+BEA)=(PBC+PBA)/ABC

PC/CF=PBC/CFB=PAC/CAF=(PBC+PAC)/(CFB+CAF)=(PBC+PAC)/ABC

Therefore

PA/AD+PB/BE+PC/CF=(PBA+PCA)/ABC+(PBC+PBA)/ABC+(PBC+PAC)/ABC

=(2PBA+2PCA+2PBC)/ABC=2(PBA+PCA+PBC)/ABC=2ABC/ABC=2