IMO Shortlist 1982 problem 9


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April 2, 2012
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Let ABC be a triangle, and let P be a point inside it such that \angle PAC = \angle PBC. The perpendiculars from P to BC and CA meet these lines at L and M, respectively, and D is the midpoint of AB. Prove that DL = DM.
Source: Međunarodna matematička olimpijada, shortlist 1982