Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 797.
(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 FCEwhich means ABE and ECF are similar. Since BC//AD, so angle FEC = angle FAD & angle FCE = angle FDAwhich means ECF and ADF are similar. Combining (1) & (2), they are isosceles and similar.
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