Wednesday, October 5, 2016

Geometry Problem 1272 Isosceles Triangle, Median, Midpoint, Perpendicular, 90 Degrees, Angles, Congruence

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

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Geometry Problem 1272 Isosceles Triangle, Median, Midpoint, Perpendicular, 90 Degrees, Angles, Congruence.

6 comments:

  1. https://goo.gl/photos/bqyrjE7bstYM7kCDA

    Draw 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 degrees
    In trapezoid FMDN , since MD ⊥ FM and E is the midpoint of NF so EM=ED
    E is the center of cyclic quadrilateral AMDB => ∠BAD=∠BMD= ∠MCN

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  2. Problem 1272
    Is <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

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  3. Area of Tr. MNC = ½ MD.NC = ½ MN.MC…(1)

    But AN = NC, NB =MN and AM = MC

    So from (1), MD.AN = NB.AM
    Hence 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 Peiris
    Moratuwa
    Sri Lanka

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  4. 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

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  5. Sumith

    Thanks for your comment.
    Peter

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  6. 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.

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