Proposed Problem
Click the figure below to see the complete problem 388.
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Complete Problem 388
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Saturday, November 14, 2009
Problem 388. Triangle, Angle bisector, Perpendicular, Parallel lines
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Join AF.
ReplyDeleteAFDG 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.
To Ramprakash.K:
ReplyDeleteThe notation of your solution (capital letters) seems to be different from the original problem.
i am sorry.
ReplyDeletechange A's to B's and B's to A's
Some conclusions.......
ReplyDelete1) BG = DE (since BEDG is an isosceles trapezoid)
2) Also EG = BD
3) < ABE = < DBF = (B-A)/2
4) < FBE = < DBG = A/2