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DFG = FCG

Source: KJMO 2023 P2

November 4, 2023
geometrycircumcircle

Problem Statement

Quadrilateral ABCD(AD‾<BC‾)ABCD (\overline{AD} < \overline{BC}) is inscribed in a circle, and H(≠A,B)H(\neq A, B) is a point on segment AB.AB. The circumcircle of triangle BCHBCH meets BDBD at E(≠B)E(\neq B) and line HEHE meets ADAD at FF. The circle passes through CC and tangent to line BDBD at EE meets EFEF at G(≠E).G(\neq E). Prove that ∠DFG=∠FCG.\angle DFG = \angle FCG.
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