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Let \(ABCD\) be a parallelogram. Let the bisector of angle \(BAD\) intersect the line \(CD\) at \(F\) and the line \(BC\) at \(G\). Prove that the circumcentre of triangle \(CFG\) lies on the circumcircle of triangle \(BCD\)

Note by Sharky Kesa 1 year, 11 months ago

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