Saturday, December 26, 2009

Problem 409: Quadrilateral, Diagonals, Perpendicular, Congruence

Proposed Problem
Click the figure below to see the complete problem 409 about Quadrilateral, Diagonals, Perpendicular, Congruence.

Problem 409: Quadrilateral, Diagonals, Perpendicular, Congruence.
See also:
Complete Problem 409
Level: High School, SAT Prep, College geometry

15 comments:

  1. there is not enough data to solve this^probleme. Values of angles ACB et ACD are irrelevant.

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  2. Values of angles ACB and CAD are relevant.

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  3. OD=a
    OA=atan80
    OB=atan80tan20
    OC=atan80tan20tan40
    tanx=OC/OD=tan20tan40tan80=tan60
    x=60
    .-.

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  4. Can this problem be solved without trig?

    bjhvash44@sbcglobal.net

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  5. Yes, this problem can be solved without trigonometry.

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  6. A synthetic Proof:

    Let P be the reflection of C over BD.
    Then the problem turns into Trigonometric Form of Ceva.
    The angles from A in clockwise are: 10, 20, 30, 40,x, 80-x.
    By a synthetic proof or as a result of trigonometric form of Ceva, 10,30,x are interchangeable; and 20,40,80-x are interchangeable.
    Swap 20 and 40.
    Then problem goes into 10,40,30,20,x,80-x.
    In that case A=10+40=50 and B=30+20=50 and C=x+80-x=80.
    Let CQ be the symmetry axis of isosceles triangle where Q is on BP.
    AQP=PQC=60 and QAP=PAC=10 => P is the incenter of triangle QAC.
    So PCA=PCQ=20 and PCB=x=60.

    While solving this kind of problems, I always swap appropriate angles. It is a powerful tool also for synthetic solutions.

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  7. Geometric solution :http://i.imgur.com/byc89Ao.jpg?1
    Let E a point ,where triangle AED is equilateral
    Triangle EBD is isosceles since ∠EFA=90 and EF=FD so ∠DEC=20
    Points E,B ,C are collinear since ∠BCA +∠BAC=∠EBA
    EA=EC since∠ EAC =∠ECA
    So Triangle DEC is isosceles so 20+40+2x=180 finally x=60

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    Replies
    1. I forget say : F is point of intersection of ED and BA
      And the point E is above AD

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    2. Excellent solution!

      Slight modification.

      Let DP be an altitude of Tr. ADB. Let DP and CB meet at E. Then Tr.s BEP and BDP are congruent ASA hence PE = PD and so Tr. ADE is equilateral.

      Now EA = EC = ED so < EDC = 80 and < EDB = 20 so x = 60

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  8. If O is the circumcentre of triangle ABD and AO extended meets BD at E, triangle OBD is equilateral. It's easy to show that triangle DOE can be rotated clockwise by 60 degrees to coincide with triangle DBC. Hence x is 60 degrees.

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  9. Referring to your comment " It's easy to show that triangle DOE can be rotated clockwise by 60 degrees to coincide with triangle DBC. Hence x is 60 degrees. ", Can you explain further .

    Peter Tran

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  10. If u can draw it with the given data u can solve it at least using trigonometry

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  11. Construct the equilateral triangle AED, like Felipe Hernandez, then <EAC=50=<ACE, so E is the circumcenter of tr. ADC and <ACD=<AED/2=30, i.e. <BDC=90-30=60.

    Best regards

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  12. Problem 409

    Let the point O is the center of the circle of des triangle ABD.Then triangle OBD is equilateral
    And <OAB=<OBA=10,<OAD=<ODA=20,<OAC=10. I take the point K on the side AC such that
    <DKC=40,so <KDO=10=<KAO , then AOKD is cyclic,then <KOD=<KAD.Therefore KO=KD (OB=BD) so BK is perpendicular bisector to OD.Then <DKB=<BKO=90+10=100,so
    <CKB=100-40=60. But <DKC=40=<DBC so DKBC is cyclic.Therefore < CDB=<CKB=60.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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  13. Problem 409
    I take the point A symmetry of A on the AB, then triangle AED is equilateral (AE=ED=AD)and <EBA=<ABD=70, but <EBA+<ABD+<DBC=70+70+40=180.So E,B and C are collinear.
    Is <EAC=<ECA=50 then AE=EC .So the point E is the center of the circumcircle of the triangle CAD.Then <ACD=(<AED)/2=60/2=30. Therefore <BDC=90-30=60.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete