Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 797.
Thursday, August 23, 2012
Problem 797: Parallelogram, Angle Bisector, Isosceles and Similar Triangles
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(1)
ReplyDeleteSince 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.
Angle BAE = angle FAB
ReplyDeleteAngle 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
Since AE bisector of BAD then ABE is half of a rhombus,
ReplyDeleteso and two others