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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to view more details of problem 1166.
Take X - reflection of E in F, AECX is a parallelogram, so AX||CE, i.e. AX_|_BE, A-X-D collinear, XED is right-angled triangle, thus EF=FX=DF, done.Best regards
Excellent work Stan!With your permission a slightly modified approach...,,,,,Let EF meet AD at X then Tr.s AXF & CEX are congruent ASASo XF = FE and so F is the centre of right Tr. XDE and hence DF = FE
Corollary: Let BF(extended) meet the circles at X,Y.Then XY is bisected at F.
Call other intersection of the 2 circles B' which is the foot of altitude from B.Let EF and AD meet at P. Then <PAF=<DBB'=<ECB', which means that AP||EC.From congruent triangles APF and CEF, PF=FE, which makes F circumcenter of right triangle PDE.