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

Labels:
angle bisector,
parallel,
perpendicular,
triangle

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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