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

Labels:
angle,
angle bisector,
quadrilateral,
triangle

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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