Saturday, April 18, 2009

Problem 282: Right Triangle, Cevian, Angles, Perpendicular, Congruence

Proposed Problem

Problem 282: Right Triangle, Cevian, Angles, Perpendicular, Congruence.

See also: Complete Problem 282, Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 8:02 AM
Labels: , , , ,

7 comments:

  1. AjitApril 21, 2009 at 10:56 PM

    In Tr DAE, a=AD*sin(2α)=2AD*sin(α)*cos(α)---(1)
    Tr BAF gives,sin(α)= x/AB and from Tr.BAD, cos(α)= AB/AD. Putting these values in (1), we've:
    a=2*AD*x/AB*AB/AD or a =2x or x=a/2
    Ajit: ajitathle@gmail.com

    ReplyDelete
  2. AnonymousNovember 21, 2009 at 12:23 PM

    draw BO, O midpoint of AD, => BO = OA = OD as radius of circle through A,B,C (B=90)
    angle FBD = angle BAD =@,(perpendicular sides)=>angle BOD = 2@ ( tr. BOD has BO=DO )
    => tr BFO and ADE are similar
    => x/BO = a/AD, but AD=2r, BO=r
    => x/r = a/2r, => x = a/2

    P.S. angle @ mean alpha

    ReplyDelete
  3. Sumith PeirisSeptember 21, 2015 at 11:51 PM

    Let ths diagonals of the cyclic quad ABDE meet at G and let H be ths midpoint of GE

    Also < GBD = < DAE so BF bisects < DBG and F is the midpoint of GD. So from midpoint theorem FH = a/2. Also FH is // to DE hence < FHB = < GDE = < BAD = < DBF = < FBG showing that Tr. BFH is isoceles

    Hence x = FH = a/2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sumith PeirisSeptember 22, 2015 at 12:08 AM

    Observe that AD is the diameter of cyclic quad ABDE and AB is the diameter of cyclic quad ABFH

    Further ABJE is also cyclic

    If BF meets AH at J, then Tr. s AGE, BGD, BFH, HFJ, ABJ are all isoceles

    ReplyDelete
  5. APOSTOLIS MANOLOUDISAugust 12, 2016 at 1:46 PM

    Problem 282
    Consider that BE intersect Ad in N and K is symmetric of B with respect to AD.Then ABDE is cyclic <DBE=2α so <EBF=α.Therefore BDKN is rhombus (ΒΝ=//ΔΚ ) and <AKD=90 .Then
    ABDK is cyclic.But <AEB=<ADB=<BND=<ANE so AN=AE.Is triangleADK=triangleAKM (DK intersect AC in M).Now DK=KM and BK=//KM so NMKB is parallelogram and 2.BF=BK=NM.
    But BK is perpendicular in AD,NM is perpendicular in AD .Now AD=AM(<ADM=<ANE=<AEN=<AMD ).So DE=MN=BK=2BF.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  6. ArifMarch 13, 2018 at 8:11 AM

    ABDE lies on a circle. Find the midpoint O of the line AD.

    O is the center of the cirlce throught ABDE.
    From O draw a line perpendicular onto the line DE. Call the line X.
    Than we have :

    DX =a/2

    Join OB. Then the triangles OXD and OBF are congruent and we have

    BF=XD
    x=a/2

    ReplyDelete
  7. Sailendra TMarch 5, 2021 at 12:49 PM

    Extend BF to meet the angle bisector of DAE at G and forms the kite DBAG.
    Easy to see DBAEG is concyclic and BD=DG=GE.
    BDGE is an isosceles trapezoid and hence DE=BG=>a=2x
    (OR)
    BDG and DGE are congruent isosceles triangles=> DE=BG

    ReplyDelete

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