Wednesday, December 12, 2012

Problem 830: Quadrilateral, Triangle, Angles, 30 degrees, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 830.

Online Geometry Problem 830: Quadrilateral, Triangle, Angles, 30 degrees, Congruence

5 comments:

  1. http://img577.imageshack.us/img577/4899/problem830.png

    See drawing per attached sketch
    Extend BC and from D draw DF ⊥to BC
    We have ∠BDF=60 and DF=1/2. BD
    In isosceles triangle ACD draw altitude CE
    We have DE=1/2.AD=1/2.BD=DF
    CE and CF are tangent lines from point C to circle centered D with radius=DE=DF ( see Sketch)
    ∠EDF=60+34=94
    ∠FCD=90-1/2. 94=43
    So in triangle BCD, external angle FCD=30+x => x=43-30=13

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  2. Geometric Solution of the problem 830 by Michael Tsourakakis from Greece
    See the image:http://img820.imageshack.us/img820/2807/geogebra.png
    solution
    CM is perpendicular of AD. Because CA=CD , CM is mediator of the AD.
    AB and MC meet of G. So triangle AGD is isosceles .So , angle GDA=73
    E is point of the GM with AED is equilateral triangle.
    Angle GDE= angle EAG =73-60=13, therefore , angle EDB=73-34-13=26
    Additional, angle DAB=73 and angle BDA=34 ,so , angle ABD=73,therefore DA=DB ,so ,DB=DE and triangle BDE is isosceles.Therefore,2 angle DBE=1800-260=1540 so, angle DBE= angle DEB=77.But angle CBD=30, therefore, angle EBC=47
    But, angle BEC=77- angle CED=77-30=47.Therefore , triangle BCE is isosceles and CB=CE .But DB=DE .So , DC is mediator of the BE ,therefore, DE is bisector of the angle BDE=26.So,x = angle CDB=13

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  3. Draw CF and DE perpendicular to AD and BCrespectively, Then easily FD equal to DE and CFD and CED Tr.s are congruent.Hence < CDE = x + 34. so 2x + 34=60. So x=13.

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Problem 830
    Form the equilateral triangle ADE ( the point C is inside the triangle ADE).So EC is bisector
    Perpendicular the AD and <AEC=<CED=30(AC=CD, AE=ED) is <BDE=<ADE-<ADB=60-34=26.
    But AD=BD=AE=ED(<ABD=180-73-34=73) then the point D is circumcenter the triangle ABE.
    So <AEB=(<ADB)/2=34/2=17 and <BAE=(<BDE)/2=26/2=13. Then <EBC=<EBA-<CBA=(180-17-13)-(73+30)=47=<BEC (=17+30=47) therefore BC=CE.But BD=DE so triangleBCD=triangleECD.
    Therefore x=<BDC=(<BDE)/2=26/2=13.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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