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 3 of 7: Equal lengths around the midpoint MM
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In plain words

The midpoint of segment CFCF between an excenter and incenter is always the midpoint of the corresponding arc, and also the center of the circle through the two feet and the two centers.

△AFE≅△FBM,AE=EF=FM=MB\triangle AFE\cong\triangle FBM,\quad AE=EF=FM=MB
Detailed analysis

Because ∠BFC=∠FBA+∠FAB=∠FAE\angle BFC=\angle FBA+\angle FAB=\angle FAE, the right-triangle angle data give △AFE≅△FBM\triangle AFE\cong\triangle FBM. Consequently AE=EF=FM=MBAE=EF=FM=MB.