Sunday, July 6, 2014

Geometry Problem 1025: Triangle, Double, Triple, Angle, Interior Point

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

Click the figure below to see the complete problem 1025.

Online Math: Geometry Problem 1025: Triangle, Double, Triple, Angle, Interior Point

2 comments:

  1. This problem may be solved in the next way:
    1.E be a point on AD such as DE = DC -> Angles: DEC = DCE = 2x
    2.Easy to see that DEC is congruent to BDC (SAS) -> EC = BC
    3.Angles: ECA = EAC = x = 3x - 2x -> EC = BC = AE
    4.Angles: BFE = 180 - BDE - BAD - EBD = 180 - 2x - 8x - (90 + 4x) = 90 -6x
    5.Let CG be perpendicular to EB thus EG = BG (Triangle EBC is isoceles)
    6.Let EF be perpendicular to AB
    7.Triangle AFE ~= BCG ~= GCE (ASA) From simple look we get that in triangle BFE: EF = EG = BG -> Angle
    BFE = 30 = 90 - 6x (As we proved above)
    Finally x = 10.

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  2. Let AD meet BC at E and CD meet AB at F. ACEF is clearly cyclic from which we can easily show FE bisects < BED = 6x and FD bisects < ADB = 8x.

    So F is the excentre of Tr. BDE and so FB bisects the exterior < . So < ABC must be 8x and solving for x we get 8x + 2x + 5x + 3x = 180 from Tr. ABC and so x = 10

    Sumith Peiris
    Moratuwa
    Sri Lanka

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