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 5 of 7: A symmetric trapezoid around MM
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In plain words

Parallel lines from the arc-midpoint structure identify exactly where the parallelogram's fourth vertex sits on the known circle.

EF∥MD,EF=FM=MD=DEEF\parallel MD,\quad EF=FM=MD=DE
Detailed analysis

From the preceding parallelogram and the equal lengths, EF∥MDEF\parallel MD and EF=FM=MD=DEEF=FM=MD=DE. Hence EFMDEFMD is a rhombus. The same angle chase gives ∠MBF=∠MFB=2α\angle MBF=\angle MFB=2\alpha and ∠MXD=∠MDX=2α\angle MXD=\angle MDX=2\alpha; with MB=MF=MD=MXMB=MF=MD=MX, quadrilateral BFDXBFDX is an isosceles trapezoid.