Sunday, January 8, 2017

Geometry Problem 1305 Triangle, Circumcircle, Angle Bisector, Arc, Perpendicular, Area

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 1305 Triangle, Circumcircle, Angle Bisector, Arc, Perpendicular, Area.

Posted by Antonio Gutierrez at 7:01 AM
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6 comments:

  1. Pradyumna AgasheJanuary 8, 2017 at 10:30 AM

    Triangle ADB and ACE are similar by AA similarity.
    We have AD/AB=AC/AE or AD.AE = AB.AC .

    Extend DH to point D' such that DH=D'H. We get AD'=AD and <DAD'= <A.
    Since H is mid-point of DD', Area of Tr. ADH = Area of Tr. AD'H, also Area of Tr. EDH = Area of Tr. ED'H . Hence Area of Tr. AEH = 1/2 Area of AD'E.

    Area of Tr. AD'E = 1/2.AD'.AE. sin(A) = 1/2.AD.AE.sin(A) = 1/2.AB.AC.sin(A)=Area of Tr. ABC. We have the result, Area of Tr. AEH = 1/2.Area of Tr.ABC


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  2. APOSTOLIS MANOLOUDISJanuary 8, 2017 at 11:20 AM

    Problem 1305
    Let AF, DG are perpendicular to BK, AB respectively.Then <AGD=<AHD=<AFD so the points A,G,D,F and H are concyclic .Hence <GFB=<GFD=<GAD=<BAE=<CAE=<CBE, so
    GF//BE. Similar HF//EC.If the BC intersects GE and HE at K,L respectively .
    Let area triangle ABC is (ABC) then (BGK)=(FKE) and ((HLC)=(FLE).
    (ABC)=(AGKLH)+(BGK)+(HLC)=(AGKLH)+(FKE)+(FLE)=(AGEH)=2.(AEH).
    APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

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  3. Peter TranJanuary 8, 2017 at 5:11 PM

    https://goo.gl/photos/2Wk2e5vJSCUmWPte8

    Draw DF ⊥ AB
    Triangle AHE congruent to AFE ( case SAS) ( see sketch)
    FH⊥AE=> Area(AFEH)=AE.MH…= 2.Area(AHE)..(1)
    Applying Ptolemy’s theorem in quad. ABEC give us AE= (b+c).BE/BC
    Note that BE/BC= AD/(2.AH) and AD.HM=AH.DH ( relation in right triangle AHD )
    Replace these values to (1) we will have Area(AFEH)=1/2. DH(b+c)
    Now Area(ABC)=Area(ADB)+Area(ADC)= ½.DH(b+c)
    Comparing these 2 areas we will have Area(AHE)= ½.Area(ABC)

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  4. Sumith PeirisJanuary 10, 2017 at 12:57 AM

    Let DH = h, altitude of ∆AHD from H be p, AE = q and BE = CE = d.

    From Ptolemy qa = d(b+c) ….(1)
    S(ADC) = ½ hb = b/(b+c).S(ABC)…(2)
    S(AHE) = ½ pq….(3)
    p/h = ½ a/d …(4) from similar ∆s.

    (3)/(2) => S(AHE) /S(ABC). (b+c)/b = pq/(hb) = ½ qa/(bd) from (4)

    Now substituting for qa/d from (1),

    S(AHE)/S(ABC) . (b+c)/b = ½ (b+c)/b

    Hence S(AHE) = ½ S(ABC)

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  5. AnonymousJanuary 12, 2017 at 11:36 AM

    Join BE and form the triangle ABE
    m(AEB) = m(ACB) (angles in same segment) --------(1)
    m(BAE) = m(EAC) = x (say) ----------(2) (given)
    So triangles AEB and ACD are similar
    => AB/AE = AD/AC
    => AB.AC = AE.AD ----------(3)
    Area of triangle AEH = 0.5*AE*AH*sin(x)
    => AEH = 0.5*AE*AD*cos(x)*sin(x)
    => AEH = 0.5*0.5*AE*AD*sin(2x)
    => AEH = 0.5*0.5*AB*AC*sin(2x) (from (3))
    => AEH = 0.5*(Area of ABC)

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  6. AnonymousJanuary 15, 2017 at 9:32 AM

    through AE, reflect HE,get H2E, H2 on AB;connect DH2;
    area AHEH2 = 2x area AHE
    connect HH2, AE perpendicular to HH2;
    Through AO make diameter AK of circle O;
    connect KH2, DK,HK,
    also connect KB,KC,KE,
    KB//DH2; KC//DH; KE//HH2;
    area BDH2 =area DKH2, (KB//DH2); area DHC= area DHK, (KC//DH);
    area KHH2 = area EHH2, (KE//HH2)

    so area ABC =area AHKH2 =area AHEH2 =2x area AHE

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