MathLabs

Problem 1

In convex pentagon ABCDEABCDE with ∠B>90∘\angle B>90^\circ, let FF be a point on segment ACAC such that ∠FBC=90∘\angle FBC=90^\circ. It is given that FA=FBFA=FB, DA=DCDA=DC, EA=EDEA=ED, and rays ACAC and ADAD trisect angle ∠BAE\angle BAE. Let MM be the midpoint of CFCF. Let XX be the point such that AMXEAMXE is a parallelogram. Show that lines FXFX, EMEM, BDBD are concurrent.
Step 2 of 7: Parallel lines and the cyclic quadrilateral BCDFBCDF
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In plain words

An angle bisector, a right angle, and a pair of similar isosceles triangles are exactly the classical fingerprint of an incenter/excenter pair.

F=incenter⁡(△DAB),C=A-excenter(△DAB),DA=DBF=\operatorname{incenter}(\triangle DAB),\quad C=A\text{-excenter}(\triangle DAB),\quad DA=DB
Detailed analysis

All six angles equal α\alpha imply ∠DAC=∠DCA\angle DAC=\angle DCA, so AB∥CDAB\parallel CD. Let G=AB∩CDG=AB\cap CD. Then ∠FDC=∠FGA=90∘\angle FDC=\angle FGA=90^\circ, while ∠FBC=90∘\angle FBC=90^\circ; hence B,C,D,FB,C,D,F are concyclic.