Tuesday, November 3, 2009

Problem 383. Concave Quadrilateral, Angle Bisectors, Semi-difference of Angles

Proposed Problem
Click the figure below to see the complete problem 383 about Concave Quadrilateral, Angle Bisectors, Semi-difference of Angles.

 Problem 383. Concave Quadrilateral, Angle Bisectors, Semi-difference of Angles.
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Complete Geometry Problem 383
Level: High School, SAT Prep

3 comments:

  1. Extend DC to met AE in F. Ext. Ang.DFE = A/2 + β and C/2 = A/2 + β + x. Also, A/2 + α = x + C/2 or A/2 + α = x + A/2+ β + x which gives us:
    2x = α -β or x = (α -β)/2
    Ajit

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  2. Let alpha = a and beta = b,
    angle BAF=angle DAF=z,angle BCD=2y
    Let seg. AE intersection seg.BC be point F.
    In tri ABF,
    angle ABF + angle FAB + angle AFB = 180
    angle AFB =180-(z+a)
    angle AFB = angle CFE...(vertex opposite angles)
    angle CFE = 180-(z+a)...(1)
    Quadrilateral ABCD is concave ,
    hence,2y=a+2z+b
    angle FCE = y=(a+2z+b)/2...(2)
    By (1) and (2) in tri CFE,
    x=180-[y+{180-(z+a)}]
    x=180-[{(a+2z+b)/2} + {180-(z+a)}]
    X=180-[{(a+2z+b-2z-2a)/2}+180]
    x=180-[{(b-a)/2}+180]
    x=180-[(b-a)/2]-180
    x=(a-b)/2

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  3. [For easy typing, I use a for alpha & b for beta]

    Let the intersection pt. of AE & BC be F

    Consider triangle BAF
    <BFE=a+<BAD/2 (ext. < of triangle)
    <EFC=180-<BFE (adj. <s on st. line)
    <EFC=180-a-<BAD/2

    Consider triangle EFC
    x=180-<EFC-<BCD/2
    =180-(180-a-<BAD/2)-<BCD/2
    =a+<BAD/2-<BCD/2----------(1)

    Consider quad. AFCD
    b=360-<FAD-reflex<BCD-<AFC
    =360-<BAD/2-(360-<BCD)-(a+<BAD/2)
    =<BCD-<BAD-a
    So, a+b=<BCD-<BAD----------(2)

    Sub (2) in (1)
    x=a-(a+b)/2
    =(a-b)/2

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