Tuesday, January 3, 2017

Geometry Problem 1302: Triangle, Median, Circles, Midpoint, Congruence

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

Details: Click on the figure below.

Geometry Problem 1302: Triangle, Median, Circles, Midpoint, Congruence.

Posted by Antonio Gutierrez at 2:51 PM
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8 comments:

  1. AnonymousJanuary 3, 2017 at 5:32 PM

    connect AE, connect CF, AE//CF;
    extend EM, extend CF, cross at G,
    <CAE=<ACF =<ACG, <AME=<GMC, AM=MC
    triangle AEM = triangleGMC,
    so GM=ME, triangle EGF = right angle, so EM=MF

    ReplyDelete
  2. Peter TranJanuary 3, 2017 at 7:09 PM

    https://goo.gl/photos/9ENrkK4G1b2hs1BCA

    Let G and H are the midpoints of AD and DC
    We have AM=GH= ½.AC
    GM= AM-GA=GH-GD= DH
    So GE=GD=MH
    And GM=DH=FH
    Triangles EGD and DHF are isosceles => ∠EGD=∠DHF
    Triangles GME congruent to HFM ( case SAS)=> ME=MF

    ReplyDelete
  3. APOSTOLIS MANOLOUDISJanuary 3, 2017 at 10:07 PM

    Problem 1302
    Is <AED=90=<DFC then AE//CF or AEEF is trapezoid.Let N midpoint in EF, so MN//AE//CF then MN is perpendicular bisector EF.Therefore ME=MF.
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  4. Sumith PeirisJanuary 3, 2017 at 11:45 PM

    Solution 1

    Let the circles be P (larger) and Q (smaller) with radii p and q respectively.

    So AM = p+q and hence PM = q and QM = p.

    Therefore easily ∆MPE ≡ ∆FQM (SAS) and hence,

    ME = MF

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. Sumith PeirisJanuary 3, 2017 at 11:47 PM

    Solution 2

    Draw MN // AE//FC, N on FE

    Let EN = u, ND = v and DF = w

    From similar ∆s,

    u/v = (p+q)/(p-q) and w/v = 2q/(p-q)

    Subtracting, u/v – w/v = (p+q-2q)/(p-q) = 1

    Hence u-w =v and so EN = NF implying that ME = MF

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  6. Sumith PeirisJanuary 4, 2017 at 12:07 AM

    Solution 3

    Let FM extended meet AE at L

    ∆MAL ≡ ∆MFC (ASA)

    So LM = MF = ME since < LEF = 90

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Stan FulgerJanuary 4, 2017 at 1:00 AM

    See that EF is the projection of AC onto BD, wherefrom the solution: the midpoint of the segment is equally apart of the ends of the projection.

    Best regards

    ReplyDelete
  8. Pradyumna AgasheJanuary 4, 2017 at 4:03 AM

    AE and CF are perpendicular to BD, Since M is midpoint of AC it lies on perpendicular bisector of EF. Hence ME=MF

    ReplyDelete

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