Ajit Athle submitted this problem, as it neatly fits in our series on Equilateral Triangles. He said:
"Here's a simple problem to go in your current series on ET's. In equilateral triangle ABC, we've cevians BD (D on AC) & CE (E on AB) such that 2*BE = AE & 2*AD = DC. If CE & BD intersect in P, then prove, w/o using trigonometry or co-ordinate geometry, that AP is perpendicular to CE."
Thank you Ajit!
Thank you Ajit!
http://geometri-problemleri.blogspot.com/2009/12/problem-56-ve-cozumu.html
ReplyDeleteHaving given the problem I may as well suggest a short solution w/o too much construction in order not to make the cure worse than the disease: Triangles ABD and BCE are congruent (AB=BC, AD=BE and included angle=60 deg.) so angle ABD = angle BCE ---(1). Now angle DPC = (angle ABC - angle ABD) + angle BCE = angle ABC using (1) And so angle DPC = 60 deg. = angle EAD which makes quad.AEPD concyclic. ----(2). Now join E to D. Clearly, AD/AE =1/2 and angle EAD is 60 deg. hence triangle EAD is 30-60-90 with angle ADE = 90 deg. and, therefore, by (2) angle APE is also 90 deg. QED
ReplyDeleteAjit
I didn't suppose that the solution is hard to be understood, sorry
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