Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1272.
https://goo.gl/photos/bqyrjE7bstYM7kCDADraw rectangle AMNF ( see sketch)Note that E is the midpoint of AB and FN.∠NAM=∠MCN( ANC is isoceles)And ∠NAM=∠FMA( AMNF os a rectangle)So ∠FMA=∠MCN => FM//NC => ∠FMC=∠NDM= 90 degreesIn trapezoid FMDN , since MD ⊥ FM and E is the midpoint of NF so EM=EDE is the center of cyclic quadrilateral AMDB => ∠BAD=∠BMD= ∠MCN
Problem 1272Is <AMD=<CNB (they have their sides perpendicular).But AM/MD=MC/MD=NC/NM=NC/NB (AM=MC, NM=NB, triangle MCD is similar with triangle NCM).Therefore triangle AMD is similar with triangle CNB .So <MAD=<NCB.Therefore <DAB=<MCN (<BAC=<BCA ).APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
Area of Tr. MNC = ½ MD.NC = ½ MN.MC…(1)But AN = NC, NB =MN and AM = MCSo from (1), MD.AN = NB.AMHence in Tr.s ANB and AMD, AN/NB = AM/MD.Further the included angles < ANB = < AMD (since < ANM = < CNM = < CMD)So it follows that the Tr.s ANB and AMD are similar and therefore < ABM = < ADM.Hence AMDB is concyclic and so< BAD = < BMD = < CAN.Sumith PeirisMoratuwaSri Lanka
Peter's comment - "In trapezoid FMDN , since MD ⊥ FM and E is the midpoint of NF so EM=ED"This is so since the perpendicular bisector of MD passes thro' E
SumithThanks for your comment.Peter
It is known that the perpendicular from N to AD passes through the midpoint P of MD. Indeed, let E be the projection of A onto CN, <PNE=<DAE (1), but triangles MND and CAE are similar, from (1) we infer P and D are homologous points, i.e. MP=PD, making NP||BD (*). With (1) we also get <MNP=<MAD (2). From tr. MNC: MN^2=ND.CN (3), but MN=BN, so from (3) we infer BN tangent to circle (BDC), or <NBD=<BCN; with (2) and (*) we conclude <CAD=<BCN, so we are done.