Advanced Geometric Problem & Solution (Twice a week update) Now 4730 problems posted.
<Qua-202-803>, hint
Consider a triangle ABD.
Let D' on BD such that JD'=//ID , then triangles JBD' and CEF are congruent ,
<JBD'=<CEF .Hence BD//EF. Hence KL is parallel to EF and BD. ID=FC
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