Several Proofs of Ceva's Theorem by Students

數 學 教 育 第 二 十 五 期 (12/2007)Several Proofs of Ceva’s Theorem by StudentsIntroductionPOON Wai Hoi BobbySt. Paul’s CollegeIn this article, some proofs of Ceva’s Theorem by students are presented.The proofs were taken from the homework given to the students in a Geometryworkshop held in the summer 2007.Ceva’s Theorem.Let P , Q and R be points on the side BC , CAand AB of triangle ABC respectively. If AP , BQ and CR areAR BP CQconcurrent, then 1.RB PC QAARQSBPFigure 1CThe proofs of Ceva’s Theorem that appeared in most classical geometrybooks are similar to the proof below, which involved ratios of areas of triangles.The Usual Proof of Ceva’s TheoremLet S be the point of concurrency (figure 1). For simplicity, let [XYZ]denote the area of any triangle XYZ . Since the areas of triangles with sameAR [ ARS ] [ ARC]height are proportional to bases, we have .RB [ RBS ] [ RBC]By property of proportions,ARRB76[ ARC][ARS ] [ ASC ] .[ RBC][RBS ] [ CSB]

EduMath 25 (12/2007)Similarly,BP [ BSA] andPC [ ASC]CQ [ CSB] .QA [ BSA]AR BP CQ [ ASC ] [ BSA][ CSB]So, 1RB PC QA [ CSB][ ASC][ BSA].Six Different Proofs of Ceva’s Theorem by the StudentsProof 1. (by Monzon Abel, Chan Chi Fai, Chan Ka Lun, Chan Wing YanJoyce, Hui Chiu Kit, Hui Man Wa, Wendy Chow, Kan Ka Yin, Lam Ching Yan,Lam Hiu Yue, Leung Ka Hei, Lo Ching Hoi, Mok Tung Hoi, Ning Man Chit, SoTsz Chung, Tan Pak Chiu, Nicole Tin, Tsang Ting Ho, Wong Cheuk Kwan,Wong Ching, Wong See Wai, Yu Lik Hang and Anglea Wu)YAXRQSBPFigure 2CLet X and Y be points on BQ produced and CR producedrespectively such that XAY is a straight line parallel to BC (figure 2).As ΔARY ~ ΔBRC and ΔAQX ~ ΔCQB , we haveAR YA CQ BC and .RB BC QA AXAs ΔASX ~ ΔPSB and ΔASY ~ ΔPSC , we haveBP PS PCBP AX , and so .AX SA YAPC YAAR BP CQ YA AX BCHence 1RB PC QA BC YA AX.77

數 學 教 育 第 二 十 五 期 (12/2007)Proof 2. (by Kwan Chin Wai, Tai Wai Ming)XAYRQSBPCFigure 3Let X and Y be points on BQ produced and CR producedrespectively such that BY and CX are both parallel to PA (figure 3).As ΔARS ~ ΔBRY and ΔAQS ~ ΔCQX , we haveAR SA CQ CX and .RB BY QA SAAs AP // XC and ΔBYS ~ ΔXCS , we haveAR BP CQ SA BY CXHence 1RB PC QA BY CX SAProof 3. (by Kan Ka Wing)ABPPC.BS BY .SX CXRQSXBPCYFigure 478

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

數 學 教 育 第 二 十 五 期 (12/2007)ASRIHenceQJASARRBRSSJBPPCQJRICQQARSSJ1PXXB.BPPCCYYPBPXBCYPCPXYPThe last equality follows as ΔRIS~ ΔJQS .Proof 5. (by Hui Shun On)AXRSQYBPFigure 6CLet X be the point on AB such that PXis parallel to CR . LetY be the point on AC such that PY is parallel to BQ (figure 6).By the intercept theorem, we haveAR AS AQ RX PC , andRX SP QY RB BCCQ BC .YQ BPHenceAQQYPCBCARRBBPPCBPPCYQQACQQABCBP1ARRX.RXRBBPPCYQQACQYQProof 6. (by Leung Yeung Yeung)Let X and Y be points on AP produced such that BX isparallel to RC and CY is parallel to QB (figure 7).By the intercept theorem, we haveAR AS andRB SXCQ SY .QA AS80

AEduMath 25 (12/2007)RQSBXPCYFigure 7As ΔBPS ~ ΔCPY and ΔBSX ~ ΔCYS , we haveBP SB SX .PC CY SYAR BP CQ AS SX SYHence 1RB PC QA SX SY AS.RemarksThe “Usual Proof” is adopted from Geometry Revisited by Coxeter andGreitzer (1967). Some of the students’ proofs appear also in some geometrytextbooks. For example, proofs similar to “Proof 1” and “Proof 2” are alsopresented in Posamentier’s book Advanced Euclidean Geometry (2002).One common point of all the proofs by the students is the use of auxiliaryparallel lines on the triangle to relate ratios of segments to other ratios ofsegments. During the workshop, I encouraged the students to first assumesome numerical values on the ratios AR : RB and CQ : QA and try tocalculate the remaining ratio BP : PC with the help of adding auxiliary81

數 學 教 育 第 二 十 五 期 (12/2007)parallel lines. By this concrete computation, students were able to understandthe relationships between ratios of the line segments on the triangle. Theproofs then should be followed by generalizing AR : RB and CQ : QA toany arbitrary ratios.The aim of this homework was to encourage students’ participation inproving geometric theorems rather than memorizing proofs. I hoped that thestudents could develop some generic problem solving skills for future geometricproblems.ReferenceCoxeter, H.S.M., Greitzer, S. L. (1967). Geometry Revisited. Mathematical Association ofAmerica.Posamentier, Alfred S. (2002).Advanced Euclidean Geometry. Key College Publishing.Email: whp@stpauls.edu.hk82