Thursday, October 6, 2016

Geometry Problem 1273 Circle, Tangent, Secant, Midpoint, Isogonal Lines, Congruent Angles

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

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Geometry Problem 1273 Circle, Tangent, Secant, Midpoint, Isogonal Lines, Congruent Angles.

7 comments:

  1. https://goo.gl/photos/bYZ9Rdpg2VBXLXcB8

    Using pole and polar properties:
    Let DE and DM cut circle O at F and G,
    Let MA cut circle O at H
    We have OM. OA= OC^2=OD^2
    So A is the pole of BC w.r.t circle O
    And A and E are harmonic conjugate points of D and F or ( A,E, D,F)= -1
    Since ∠EMA= 90 so Apollonius circle diameter AE will pass through M
    So ME is a angle bisector of angle MDF.
    And ∠AMD=∠FMH=∠HMG => BC//GF
    So Arc BG=Arc CF and ∠BDM=∠EDC

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  2. https://goo.gl/photos/bYZ9Rdpg2VBXLXcB8

    Using elementary geometry:
    Let DE and DM cut circle O at F and G,
    Let MA cut circle O at H
    We have OM. OA= OC^2=OD^2
    So OA/OF=OF/OM => triangle MOF similar to triangle FOA ( case SAS)
    And FA/FO=FM/MO….. (1)
    Similarly triangle MOD similar to triangle DOA ( case SAS)
    And DA/DO=MD/MO ….. (2)
    Divide (1) to (2) and note that OD=OF
    We have FA/DA=FM/MD => MA is a external angle bisector of angle DMF
    And ∠AMD=∠FMH=∠HMG => BC//GF
    So Arc BG= Arc CF and angle BDM= angle CDE

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  3. Problem 1273
    If AD and DM cut circle O at K and L.Then AB^2=AM.AO=AC^2=AD.AK. So MOKD is cyclic.
    OD^2=OB^2=OM.OA so triangle ADO is similar with triangle DMO.Therefore <OMD=<ODA
    or <AMD=<ODK=<OMK. So <KMC=<CMD =<KMD/2=<KOD/2=<KBD, then <CMD=<CBD+<BDL=<KMC=<CBD+<CBK. So <BDL=<KBC=<KDC.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  4. Reading thro the 2nd proof of Peter, give below a slightly modified approach

    Let AD meet the circle at P.
    Let < MAP = p, < PAC = q, < ACD = < DBC = < DPC = r and < ODP = < OPD = s

    Then OC^2 = OM. OA = OD^2 = OP^2
    So < MDO = < MPA = p hence OMDP is concyclic

    Therefore < AMD = s and so
    < BDM = 90-s-r .....(1)

    Now < DOC = 2r so < CDE = 90-r-s (since < ODP = s)

    Therefore < BDM = < CDE = 90-s-r

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  5. How about the converse "if < BDM = < CDE then M is the mid point"?

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  6. The symmedian passes through the intersection of the tangents to the circumcircle at the extremities of the opposite side, i.e. AD is symmedian of tr. DBC.

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  7. Extend AD to meet the circle O at X.Connect AMO and form the right triangle ABO
    In the triangle ABO
    AB^2 = AM.AO
    But AB^2 = AD.AX
    Hence D,M,O and X are concyclic points.

    Let Angle XDC = EDC = @ and ODX = $
    => Angle OXD = $ => OMD = 180-$ => AMD = $ => BMD = 90+$ -------------(1)
    Join OC and consider the isosceles triangle ODC
    We know Angle ODC = @+$ = OCD => Angle DOC = 180-2(@+$) => Angle DBC = DBM = DOC/2 = 90-(@+$)----------(2)

    Now consider the triangle BDM and from (1) and (2)
    Angle BDM = 180-(90+$)-(90-(@+$)) = @ = EDC

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