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< ACB +2 < ACD = 180^o , BC +CE = AE

Source: Norwegian Mathematical Olympiad 2003 - Abel Competition p3

February 21, 2020
anglesgeometry

Problem Statement

Let ABCABC be a triangle with AC>BCAC> BC, and let SS be the circumscribed circle of the triangle. ABAB divides SS into two arcs. Let DD be the midpoint of the arc containing CC. (a) Show that ∠ACB+2⋅∠ACD=180o\angle ACB +2 \cdot \angle ACD = 180^o. (b) Let EE be the foot of the altitude from DD on ACAC. Show that BC+CE=AEBC +CE = AE.
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