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

5 comments:

  1. 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

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  2. 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

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  3. 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

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  4. 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

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  5. 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

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