Saturday, November 14, 2009

Problem 388. Triangle, Angle bisector, Perpendicular, Parallel lines

Proposed Problem
Click the figure below to see the complete problem 388.

 Problem 388. Triangle, Angle bisector, Perpendicular, Parallel lines.
See also:
Complete Problem 388
Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

3 comments:

  1. Join AF.
    AFDG is cyclic.(AGF = ADF = 90)
    angle BAD is 90-B/2(AMS of a triangle)
    angle CAE is 90-C/2(AMS of a triangle)
    so, angle EAD is B/2+C/2 {(90-B/2)+(90-C/2)-A}
    angleADG = angleAFG = 90-A/2 = B/2+C/2 = angleEAD (angle in the same segment)
    SO, DG||AE.

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  2. To Ramprakash.K:
    The notation of your solution (capital letters) seems to be different from the original problem.

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  3. i am sorry.
    change A's to B's and B's to A's

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