Azim Premji University At Right Angles, July 2023 87
Student Corner
M
enelaus’ theorem is an extremely important result in higher Euclidean geometry. We offer a proof of the theorem here using the sine rule from trigonometry.Menelaus’s Theorem
In∆ABC a transversal line crosses the side lines CA, AB and BC at points Q, P and M, respectively. Then we have the following equality:
BM MC ·CQ
QA ·AP PB =1. Proof
Apply the sine law in∆APQ:
sinθ1
PQ = sinθ4
AP = sin(180◦−θ1−θ4)
QA ,
therefore sinθ4
AP = sin(θ1+θ4)
QA =⇒ AP
QA = sinθ4
sin(θ1+θ4).
1
Menelaus’s Theorem
ANKUSH KUMAR PARCHA
Keywords: Euclidean geometry, Menelaus, proof, sine rule.
88 Azim Premji University At Right Angles, July 2023
Again, applying the sine law in∆QCM:
sinθ3
QM = sinθ4
CM = sin(180◦+θ3−θ4)
CQ ,
therefore sinθ4
CM = sin(θ3−θ4)
CQ =⇒ CQ
CM = sin(θ3−θ4) sinθ4 . Again, applying the sine law in∆PBM:
sin(180◦+θ3−θ4)
PB = sin(θ1+θ4)
BM = sinθ2,
PM therefore sin(θ3−θ4)
PB = sin(θ1+θ4)
BM =⇒ BM
PB = sin(θ1+θ4) sin(θ3−θ4). Multiplying the corresponding sides of the three equalities, we get:
AP QA· CQ
CM· BM
PB = sinθ4
sin(θ1+θ4) ·sin(θ3−θ4)
sinθ4 ·sin(θ1+θ4) sin(θ3−θ4), therefore BM
MC ·CQ QA ·AP
PB =1. Corollary
(
1+ AP PB
) (
1+ CM CB
)
= QM QP · QA
QC. Proof
For∆MBP with AQC as transversal, Menelaus’s Theorem can be written as BA
AP · PQ QM ·MC
CB =1. Multiplying this with the previous result, we get:
BA AP · PQ
QM· MC CB · BM
MC ·CQ QA · AP
PB =1, therefore BA
PB ·BM
CB = QM QP · QA
QC =⇒ BP+PA
PB ·BC+CM
CB = QM QP · QA
QC.
Or: (
1+ AP PB
) (
1+ CM CB
)
= QM QP · QA
QC.
ANKUSH KUMAR PARCHA is currently studying BSc Physics Honours (second year) at Indira Gandhi National Open University (IGNOU), New Delhi. He has a deep interest in doing research in Physics and Mathematics. He has contributed many articles, problems, and solutions to magazines such as Mathematical Gazette, AMJ, ISROSET, SSMJ, Pentagon, Octagon, At Right Angles and Mathematical Reflections. He may be contacted at [email protected].