Proposed Problem
Click the figure below to see the complete problem 381 about Quadrilateral, Diagonals, Angle Bisectors, Angles.
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Complete Geometry Problem 381
Level: High School, SAT Prep, College geometry
Saturday, October 31, 2009
Problem 381. Quadrilateral, Diagonals, Angle Bisectors, Angles
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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.
ReplyDeleteBesides, angle DGA=x+m. So, x+m+n=alpha+2m=beta+2n. solving, x=(alpha+beta)/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
ReplyDelete2. 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
< sum of a pentagram=180
ReplyDeleteSo, <CBD+<BCA=180-x-<EAC-<EDB
180-x-<EAC-<EDB+a+<BAC=180
a=x+<EAC+<EDB-<BAC
a=x+0.5blue-0.5green
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180-x-<EAC-<EDB+b+<CDB=180
b=x+<EAC+<EDB-<CDB
b=x+0.5green-0.5blue
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a+b=2x
x=(a+b)/2