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RU-SPB 2014 Problem 103

  • Grade 10

DD is inner point of triangle ABCABC . EE is on BDBD and CE=BDCE=BD . ∠ABD=∠ECD=10,∠BAD=40,∠CED=60\angle ABD=\angle ECD=10,\angle BAD=40,\angle CED=60 Prove, that AB>ACAB>AC

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