Thursday, November 25, 2010

Problem 541: Right Triangle, Altitude, Angle Bisector, Congruence

Geometry Problem
Click the figure below to see the complete problem 541 about Right Triangle, Altitude, Angle Bisector, Congruence.

Problem 541. Right Triangle, Altitude, Angle Bisector, Congruence.
Go to Complete Problem 541

Posted by Antonio Gutierrez at 1:31 PM
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6 comments:

  1. c .t . e. oNovember 26, 2010 at 1:05 PM

    draw EP perpendicular to AC
    ▲ADF ~ ▲AEP => AP/EP = AD/DF
    ▲ABE ~ ▲AEP => AB/BE = AP/EP
    AF bisector => AB/BF = AD/DF
    =>
    AB/BE = AB/BF
    =>
    BE = BF

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  2. PravinNovember 29, 2010 at 7:17 PM

    Consider right triangles ABE, AFD
    Complements of the equal angles BAE,FAD are equal
    So Angle BEF = Angle AFD
    = Angle BFE (vertically opposite)
    etc

    ReplyDelete
  3. Neha PrabhuneDecember 15, 2010 at 6:29 AM

    Triangle ABE ~ Triangle ADF by AA Test of similarity
    .....(angle FAD=angle BAF{given}, angle ABC=angle ADB=90{given})
    By c.a.s.t angle AFD=angle FEB
    angle AFD=angle BFE ....(vertically opposite angles)
    Hence, angle BFE=angle FEB.
    Therefore BF=BE....(Isosceles Angle Theorem)

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  4. Sumith PeirisJuly 1, 2015 at 6:38 AM

    The complements of both < BEA and AFD are A/2 and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. APOSTOLIS MANOLOUDISJune 22, 2016 at 4:28 AM

    Problem 541
    Is<BEF=<ECA+<EAC=90-<BAE=90-<A/2, but <BFE=<AFD=90-<FAD=90-<A/2=
    Therefore BE=BF.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  6. MarcoDecember 24, 2023 at 9:15 PM

    Let <BAE=<EAC=x
    <ACB=90-2x
    <FBE=<DBC=2x
    <BEF=<BEA=90-x
    <BFE=<180-<FBE-<BEF=180-2x-(90-x)=90-x-<BEF
    Since <BEF=<BFE, BF=BE

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