Thursday, August 23, 2012

Problem 797: Parallelogram, Angle Bisector, Isosceles and Similar Triangles

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 797.

Online Geometry Problem 797: Parallelogram, Angle Bisector, Isosceles and Similar Triangles.

2 comments:

  1. (1)
    Since angle BAE = angle DAE = angle BEA,
    thus BA = BE, and so ABE is isosceles.

    (2)
    Since AB//DC, so
    angle BAE = angle CFE & angle ABE = angle FCE
    which means ABE and ECF are similar.

    Since BC//AD, so
    angle FEC = angle FAD & angle FCE = angle FDA
    which means ECF and ADF are similar.

    Combining (1) & (2),
    they are isosceles and similar.

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  2. Angle BAE = angle FAB
    Angle BAE = angle AFB (parallel lines)
    Angle BEA = angle FAB ( parallel lines)
    So all these four angles are the same and hence the required triangles are isos and similar

    ReplyDelete