Monday, August 30, 2010

Problem 516: Triangle, Cevian, Congruence, Angles

Geometry Problem
Click the figure below to see the complete problem 516 about Triangle, Cevian, Congruence, Angles.

Problem 516: Triangle, Cevian, Congruence, Angles


See also:
Complete Problem 516

Level: High School, SAT Prep, College geometry

5 comments:

  1. x=11.25deg. (4x=45deg).

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  2. Let me give some supplement:

    Let point E, AE//DB and AD//EB.
    Triangle ACE is isosceles, angle EAC = angle BDC.
    So, angle AEC = angle ACE = 5x, that is, angle ECB = x.
    Angle EAB = x, since AE//DB.

    EA=BC, since points A, C, B, E are on a circle, and AC//EB.
    So, triangle ABE and CEB is congruence, that is,
    CB = EA = BD, so triangle DCB is isosceles ---> angle BDC is 6x.
    6x + 6x + 4x = 16x = 180deg. ---> x = 11.25deg.

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  3. Draw AE//BD with AE=BD
    ADBE is a parallelogram and AEC isosceles . angle (EAC)=180-10x , angle(AEC)=5x
    AEBC is a cyclic quadrilateral with BE//AC so AEBC will become isosceles trapezoid with AE=BC
    Angle(BAC)=angle (ECA)=5x
    In triangle ABC we have 5x+5x+6x=180 so x=11.25

    Peter Tran

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  4. Mark E on AC extended such that < CBE = x. Then DB = DE = AC so AD = CE
    Now draw circle BDE to cut AB at P and BE at Q. Since equal arcs subtend equal angles in a circle PD = CQ which in turn subtend equal angles at Q and D showing that PQ//DC

    So PDCQ is an isoceles trapezoid and Tr.s APD & CQE are congruent making Tr. ABE and hence Tr. BDC both isoceles. So < BDC = 6x and therefore 6x + 6x + 4x = 180 and x = 11.25

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  5. Problem 516
    Let E symmetry of D on the AB, then EB=BD=AC and <EBA=<DBA=x and <AEB=<ADB=10x.But
    <EBC=6x=<ACB then the AEBC is isosceles trapezoid (ΑΕ//ΒC).So <AEB+<EBC=180 or 10x+6x=180 therefore x=11.25
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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