Saturday, September 3, 2016

Geometry Problem 1255: Right Triangle, Angle, 90 Degrees, Perpendicular, Collinear Points

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

Click the figure below to view more details of problem 1255.

Geometry Problem 1255 Elements: Geometry Problem 1255.

Posted by Antonio Gutierrez at 6:36 PM
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5 comments:

  1. APOSTOLIS MANOLOUDISSeptember 3, 2016 at 11:12 PM

    Problem 1255
    Solution 1
    In accordance with problem 848 in triangle BCD <BDC=26.5 so <BCD=90-26.5=63.5.
    But accordance with problem 849 in triangle CDE <DEC=18.5 so <DCE=90-18.5=71.5.
    Now <ACB+<BCD+<DCE=45+63.5+71.5=180.Therefore A,C, and E are collinear.
    Solution 2
    Let the points K,L on the side DE such that DK=KL=LE. If <DKC=α, <DLC=β and <DEC=θ
    then in accordance with problem 363 in triangle DCE apply α+β+θ=90 or β+θ=45.
    Is triangle BCD similar with triangle DCL(BC=2BD, DC=2DL) so <BCD=<DCL=90-β and
    <DCE=90-θ. But <ACB+<BCD<DCE=45+(90-β)+(90-θ)=45+180-(β+θ)=45+180-45=180.
    Therefore A,C, and E are collinear
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  2. Peter TranSeptember 4, 2016 at 12:07 AM

    Geometry solution:
    https://goo.gl/photos/sdw3M9H3Ktv2F8PVA

    Let x=AB=BC and BD=2x
    Let AC cut circumcircle of triangle BDC at F
    Note that ADF is 45-45-90 triangle so
    DF=AF=3x/√(2)
    CF=AF-AC=3x/√(2) – x√2= x/√(2)
    And CD/DE=CF/DF=1/3
    Triangle DFC similar to EDC ( case SAS)
    So angle (DCF)= angle (DCE)=> A, C and E are collinear.

    Trigonometry solution:
    Let u= ∠ (BCD) and v= ∠ (DCE)
    We have ∠ (BCE)=u+v
    So tan(∠ (BCE))= (tan(u)+tan(v))/(1- tan(u).tan(v))= (2+3)/(1-2.3)= -1
    So ∠ (BCE=135 degrees and ∠ (ACE)= 45+135= 180 degrees
    So A,C and E are collinear

    ReplyDelete
  3. Sumith PeirisSeptember 4, 2016 at 1:51 AM

    Geometry Solution

    Drop a perpendicular EF from E to AF.
    Let AB = BC = p and CD = q so that BD = 2p and DE = 3q

    Now triangles CBD and DFE are similar hence DE = 3p and FE = 6p since DE = 3DC.

    So AF = 6p = FE hence < FAE = 45. But < FAC = 45
    It thus follows that A,C,E are collinear

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  4. Sumith PeirisSeptember 4, 2016 at 1:54 AM

    Trigonometry Solution

    Let < BCD = m and let < DCE = n

    So tan(m+n) = (2+3)/(1-2X3) = -1
    Hence m+n = 135 degrees and the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  5. AnonymousSeptember 6, 2016 at 12:20 PM

    Let AB = BC = 1 => AC = Sqrt(2)
    BD = 2 => DC = Sqrt(5)

    Join CE and form the Right triangle CDE
    CD = Sqrt(5)
    DE = 3*Sqrt(5)
    CE = 5*Sqrt(2)


    Drop a _|_ from D to EC that meets at F, DF = 3/Sqrt(2)

    Area of ADC = 0.5*AD*BC = 0.5*3*1 --------(1)
    Area of DCE = 0.5*DC*DE = 0.5*Sqrt(5)*3*Sqrt(5) = 0.5*5*3 -------------(2)

    Adding (1) & (2) = 9 Sq.units ---------(3)

    Let us assume A,C & E are collinear

    Join AE and form the triangle ADE.

    Area of ADE = 0.5*AE*DF = 0.5*6*Sqrt(2)*3/Sqrt(2) = 9 Sq.units ------------(4)

    Since (3) = (4), our assumption is true and A,C & E are collinear

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