Saturday, October 31, 2009

Problem 381. Quadrilateral, Diagonals, Angle Bisectors, Angles

Proposed Problem
Click the figure below to see the complete problem 381 about Quadrilateral, Diagonals, Angle Bisectors, Angles.

 Problem 381. Quadrilateral, Diagonals, Angle Bisectors, Angles.
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Complete Geometry Problem 381
Level: High School, SAT Prep, College geometry

2 comments:

  1. Let F be the intersection of BD and AC , G be the intersection of ED and AC ,angle BAE=m and angle CDE=n. Then, angle BFC=alpha+2m=beta+2n.
    Besides, angle DGA=x+m. So, x+m+n=alpha+2m=beta+2n. solving, x=(alpha+beta)/2.

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  2. 1. Let F be the intersection of BD and AC , G be the intersection of ED and AC, angle BAE=m and angle CDE=n

    2. Angle chase DFE = 180 - (beta + n) and AGE = 180 - (alpha + m)
    3. Then consider triangles AEG and DEF:
    for AEG: x + m + (180 - (beta + n)) = 180 or x + m = beta + n
    for DEF: x + n + (180 - (alpha + m)) = 180 or x + n = alpha + m.

    4. Add the two equations together: 2x + (m+n) = alpha + beta + (m+n) and
    simplify to get x = (alpha + beta) / 2



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