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QE // BC wanted, <ACB=60^o, AE / BD = BC / CE -1, 2 circumcircles

Source: OIFMAT I 2010 day 2 p6 - Chilean Math Forum FMAT Olympiad https://artofproblemsolving.com/community/c2484778_oifmat

September 25, 2021
parallelgeometry

Problem Statement

Let △ABC \vartriangle ABC be a triangle with ∠ACB=60º \angle ACB = 60º . Let E E be a point inside AC‾ \overline {AC} such that CE<BC CE <BC . Let D D over BC‾ \overline {BC} such that AEBD=BCCE−1. \frac {AE} {BD} = \frac {BC} {CE} -1 . Let us call P P the intersection of AD‾ \overline {AD} with BE‾ \overline {BE} and Q Q the other point of intersection of the circumcircles of the triangles AEP AEP and BDP BDP . Prove that QE∥BCQE \parallel BC .
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