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 1 of 7: Naming the trisection angle
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In plain words

The trisection of ∠BAE\angle BAE combined with the two given isosceles triangles FABFAB and DACDAC produces one repeated angle throughout the figure.

α:=∠FBA=∠FAB=∠FAD=∠FCD=∠DAE=∠ADE\alpha:=\angle FBA=\angle FAB=\angle FAD=\angle FCD=\angle DAE=\angle ADE
Detailed analysis

Since rays AC,ADAC,AD trisect ∠BAE\angle BAE, and FA=FBFA=FB makes △FAB\triangle FAB isosceles while DA=DCDA=DC makes △DAC\triangle DAC isosceles, and EA=EDEA=ED makes △EAD\triangle EAD isosceles, all six angles ∠FAB,∠FBA,∠DAC,∠DCA,∠EAD,∠EDA\angle FAB,\angle FBA,\angle DAC,\angle DCA,\angle EAD,\angle EDA are equal to a common value α=13∠BAE\alpha=\tfrac13\angle BAE.