Several Proofs of Ceva's Theorem by Students

EduMath 25 (12/2007)Let X and Y be points on the line BC such that AX // BQ andAY // RC (figure 4).As ΔCQB ~ ΔCAX and ΔBRC ~ ΔBAY , we haveCQ BC AR CY and .QA BX RB BCAs ΔBPS ~ ΔXPA and ΔPSC ~ ΔPAY , we haveBP PS PCBP XB , and so .XB SA CYPC CYAR BP CQ CY BX BCHence 1RB PC QA BC CY BX.Proof 4. (by Wong Wai Lung)ARQISJBXPYCFigure 5Let X and Y be points on BC such that XR and YQ areboth parallel to PA . Let I be the point of intersection of BQ andXR , and J be the point of intersection of CR and YQ (figure 5).By the intercept theorem, we haveAR PX andRB XBCQ CY .QA YPAs XR // PA and ΔBIR ~ ΔBSA ,we havePBXBSB AS .IB RIAs YQ // PA and ΔCJQ~ ΔCSA , we haveCYCPCJ QJ .CS AS79