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Area of convex quadrilatera

Source: St Petersburg Olympiad 2011, Grade 9, P5

September 15, 2017
geometry

Problem Statement

ABCDABCD - convex quadrilateral. A+D=150,B<150,C<150\angle A+ \angle D=150, \angle B<150, \angle C<150 Prove, that area ABCDABCD is greater than 14(ABCD+ABBC+BCCD)\frac{1}{4}(AB*CD+AB*BC+BC*CD)