Problem:
Let ABC be a triangle. A circle passing through A and B intersects segments AC and BC at D and E, respectively. Lines AB and DE intersect at F while lines BD and CF intersect at M. Prove that MF=MC if and only if MBâ‹…MD=MC2.
Solution:
First Solution: Extend segment DM through M to G such that FG∥CD.
Then MF=MC if and only if quadrilateral CDFG is a parallelogram, or, FD∥CG. Hence MC=MF if and only if ∠GCD=∠FDA, that is, ∠FDA+∠CGF=180∘.
Because quadrilateral ABED is cyclic, ∠FDA=∠ABE. It follows that MC=MF if and only if
180∘=∠FDA+∠CGF=∠ABE+∠CGF
that is, quadrilateral CBFG is cyclic, which is equivalent to
∠CBM=∠CBG=∠CFG=∠DCF=∠DCM
Because ∠DMC=∠CMB,∠CBM=∠DCM if and only if triangles BCM and CDM are similar, that is
BMCM​=CMDM​
or MBâ‹…MD=MC2.
Second Solution: We first assume that MB⋅MD=MC2. Because MDMC​=MCMB​ and ∠CMD=∠BMC, triangles CMD and BMC are similar. Consequently, ∠MCD=∠MBC.
Because quadrilateral ABED is cyclic, ∠DAE=∠DBE. Hence
∠FCA=∠MCD=∠MBC=∠DBE=∠DAE=∠CAE,
implying that AE∥CF, so ∠AEF=∠CFE. Because quadrilateral ABED is cyclic, ∠ABD=∠AED. Hence
∠FBM=∠ABD=∠AED=∠AEF=∠CFE=∠MFD.
Because ∠FBM=∠DFM and ∠FMB=∠DMF, triangles BFM and FDM are similar. Consequently, DMFM​=FMBM​, or FM2=BM⋅DM=CM2. Therefore MC2=MB⋅MD implies MC=MF.
Now we assume that MC=MF. Applying Ceva's Theorem to triangle BCF and cevians BM,CA,FE gives
AFBA​⋅MCFM​⋅EBCE​=1
implying that AFBA​=ECBE​, so AE∥CF.
Consequently, ∠DCM=∠DAE. Because quadrilateral ABED is cyclic, ∠DAE=∠DBE. Hence
∠DCM=∠DAE=∠DBE=∠CBM
Because ∠CBM=∠DCM and ∠CMB=∠DMC, triangles BCM and CDM are similar. Consequently, DMCM​=CMBM​, or CM2=BM⋅DM.
Combining the above, we conclude that MF=MC if and only if MBâ‹…MD=MC2.
The problems on this page are the property of the MAA's American Mathematics Competitions
