Sunday, May 18, 2008

Elearn Geometry Problem 4



See complete Problem 4
Triangle, Quadrilateral, Angles, Congruence. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

15 comments:

  1. Trazamos el segmento BD

    El triangulo BDC es isosceles => angulo(DBC)=angulo(BDC)=(180-2A)/2= 90-A

    Tomamos AD=DC=BC=1

    En el triangulo BDC ; 1/sen(90-A)=BD/sen(2A) => BD=2 sen(A)

    En el tringulo ABD ; 1/sen(ABD)=2sen(A) / sen (A) => sen(ABD)= 1/2 => angulo (ABD) =30

    Angulo(x)=angulo(DBC)+angulo(ABD)=90-A + 30= 120 - A

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  2. It is possible to provide a proof using only elementary geometry!

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  3. Let point E be a reflection of point D over AB.
    Then AEB and ADB triangles are congruent
    Angle EAB = angle DAB = alpha => angle EAD = 2 alpha => triangles AED and CBD are congruent (AE = AD = DC = BC) => ED = DB
    But BE = DB (because AEB and ADB are congruent) => EBD is an equilateral triangle.

    Angle EBD = 60 => angle ABD = 30 (since ABD = ABE)
    Angle DBC = 90 - alpha (since DCB is an isosceles triangle)

    x = angle ABD + angle DBC = 30 + (90 - alpha) = 120 - alpha.

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  4. This comment has been removed by a blog administrator.

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  5. http://mate-facil.co.cc/35/
    By TiNo

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  6. Let the bisector of <DCB intersect AB at X. Now since <XCA = <XAD = a, AD = CD and XD is common, by SAS, AXD is congruent to XCD. Hence <DXC = <DXA. Since X is on the perpendicular bisector of BD it also follows that <DXC = <BXC. Therefore since <DXC = <DXA = <BXC and all three angles are supplementary, <BXC = 60. Hence <XBD = 30. Now, <XBC = <XBD + <DBC = 30 + 90 - a = 120 - a.

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  7. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

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  8. Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 4. Thanks Eder.

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  9. solution is uploaded to the following link

    https://docs.google.com/open?id=0B6XXCq92fLJJajJOTFRfNTlSTDZGdUlVZEJfSml4UQ

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  10. Proof:
    Join AC
    Through D draw a line perpendicular to AC and meet AC at M and meet AB at E.
    ∵ AD = DC
    ∴ EM is the perpendicular bisector of AC
    ∴ AE = EC and ∠AEM = ∠CEM
    ∴ ∠EAD = ∠ECD = α
    ∵ DC = CB
    ∠DCE = ∠BCE = α
    Δ DCE ≡ Δ BCE
    ∠CEM = ∠CEB
    ∵ ∠AEM = ∠CEM = ∠CEB = 60o
    ∵ In ΔBCE, x + α = 120o

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  11. Join BD.
    Let E be the circumcentre of ∆ABD.
    ∠DEB = 2∠DAB = 2α.
    ∆s DEB, DCB are congruent (Each isosceles, common base BD,
    equal vertical angles, each 2α)
    Follows EA = ED = DC = DA, ∆EAD is equilateral, and ∠AED = 60°
    So ∠ABD = (1/2)∠AED = 30°
    Observe that ∠DBC = 90° - α
    Hence x = 30° + (90° - α) = 120° - α

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  12. hallo my dear brothers and sisters ...... here is a simple solution for this problem
    lets start
    join A,C. join B,D.
    draw an angle bisector of angle DCB.it intersects DB at the point F and
    also intersects AB , at the point E.
    now consider triangle DBC
    CD=CB (given)
    angle DCF=angle FCB=alpha
    angle FDC = angle FBC = (180- 2 alpha)/2=90-alpha
    so angle DFC = angle BFC = 90
    so angle EFD = angle EFB = 90
    now consider two triangle DFC and BFC
    cf is common
    angle DCF=angle FCB=alpha
    angle DCF=angle FCB=alpha
    triangle CDF is congruent with triangle CBF
    so FD = FB

    now consider two triangle DEF and EFB
    EF is common
    FD = FB
    angle EFD = angle EFB = 90
    two triangle DEF and EFB are congruent with each other
    so angle BEF = angle FED............... (1)
    consider triangle ADC
    AD = DC
    this is an isosceles triangle
    so angle DAC = angle DCA
    consider triangle AEC
    angle EAC = angle EAD + angle DAC
    angle ECA = angle ECD + angle DCA
    angle EAD = angle ECD = alpha
    so angle EAC = angle ECA
    so triangle AEC is isosceles triangle
    so AE = EC
    now consider two triangle AED and CED
    AE=CE
    ED is common
    AD=DC
    triangle AED and CED are congruent
    so angle AED = angle FED ..........(2)
    from (1) and (2)
    angle BEF = angle FED = angle AED = 180/3 =60
    so angle EBF = 90 - 60 = 30
    ANGLE ABC= angle EBF + angle FBC
    = 30 +90 - alpha = 120 - alpha

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  13. Draw an angle bisector of angle BCD and let it meet at the point E on the side AB
    Then connect E and D.
    Since In Tri ACD , angle ACD = angle CAD
    angle ECD = angle EAD = alpha
    CE = AE ( Isosceles tri AEC )
    Since In Tri BCE and Tri ADE
    BC = AD ( given )
    angle BCE = angle EAD = alpha
    CE = AE (proved)
    Tri BCE congruent to Tri ADE. This implies x = angle CBE = angle ADE
    Since in Tri CDE and in Tri ADE
    CE = AE ( proved )
    CD = AD ( given )
    angle ECD = angle EAD
    Tri CED congruent Tri ADE , angle CDE = angle ADE = x
    Since Tri BCE , Tri CDE , Tri ADE are congruent to each other
    angle BEC = angle CED = angle AED
    and BEC + CED + AED =180 degree
    3(AED) = 180 degree
    AED = 60 degree
    IN Tri CEA , angle CEA = 2(AED) angle = 120 degree. Since it Is isosceles
    angle ECA = angle EAC = 60 degree
    IN Tri CDA , angle DCA = angle DAC = 60 - alpha , and angle CDA = 120 + 2(alpha)
    AND FINALLY
    ANGLE CDA = 360 - (ANGLE EDC + ANGLE EDA)
    ANGLE CDA = 360 - (ANGLE EDC + ANGLE EDC) ( EDC = EDA)
    ANGLE CDA = 360 - ANGLE 2(EDC)
    ANGLE CDA = 360 - 2X (EDC = X)
    120 + 2(ALPHA) = 360 - 2X (ANGLE CDA = 120 + 2(ALPHA))
    2X - 360 = - 2(ALPHA) - 120
    2X = 360 - 120 - 2(ALPHA)
    = 240 - 2(ALPHA)
    = 2( 120 - ALPHA)
    THEREFORE X = 120 - ALPHA

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  14. Let the bisector of < DCB meet AB at E and BD at F. Draw altitude DG of Tr. ABD.

    Tr. s BCF & DCF are congruent and these are also congruent with Tr. ADG. So DG = DF = FB hence in right Tr. BGD BD = 2DG which means that < ABD = 30 and since < DBC = 90-@ the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  15. https://www.youtube.com/watch?v=2XwhJStz4bA

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