See complete Problem 245 at:
gogeometry.com/problem/p245_parallelogram_equilateral_triangle_congruence.htm
Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Thursday, February 5, 2009
Elearn Geometry Problem 245: Parallelogram with Equilateral triangles. Congruence
Subscribe to:
Post Comments (Atom)
BE=AD=EC ; AB=DF=CF ; angle ABE = angle ADF = angle ECF , so the triangles ABE , FDA , FCE are congruents, so AE=AF=EF.
ReplyDeleteSo what if i want to find the value of BEC or BCE angle?
ReplyDeleteDF=DC=AB
ReplyDeleteAF=AE
AD=BC=BE
Triangle DAF congruent to triangle BEA
So <BEA=<DAF=x
Let AE and BC meets at G
<BGA=60+x (ext < of triangle)
<EAD=<BGA=60+x (alt <s, BC//AD)
<EAF=<EAD-<DAF=60
---------------------------------------
DF=FC
DA=BC=CE
FA=FE
Triangle DAF congruent to triangle CEF
So, <DFA=<CFE
<CFD=<DFA+<CFA=60
<CFA=60-<DFA=60-<CFE
So, <CFE+<CFA=60=EFA
So, AEF is an equilateral triangle since all 3 <s=60
Triangles ABE, ECF, ADF are congruent SAS
ReplyDeleteSo AE = EF = FA
Hence Triangle AEF is equilateral
Sumith Peiris
Moratuwa
Sri Lanka