Problem 1

In convex pentagon

ABCDE

with

\angle B>90^\circ

, let

F

be a point on segment

AC

such that

\angle FBC=90^\circ

. It is given that

FA=FB

DA=DC

EA=ED

, and rays

AC

and

AD

trisect angle

\angle BAE

. Let

M

be the midpoint of

CF

. Let

X

be the point such that

AMXE

is a parallelogram. Show that lines

FX

EM

BD

are concurrent.

Step 7 of 7: Radical axis theorem finishes the proof

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In plain words

Three circles pairwise sharing the lines $BD$ , $FX$ , $EM$ as their radical axes must have those three lines concurrent.

\text{radical axes of } (AEDMB),\,(BFDXC),\,(EXMF)\ \text{concur}

Detailed analysis

Consider the three circles $(AEDMB)$ (through $A,E,D,M,B$ , all established to be concyclic in the steps above), $(BFDXC)$ (through $B,F,D,X,C$ , i.e. circle $(BFDC)$ from Step 3 together with $X$ from Step 4), and $(EXMF)$ (the isosceles trapezoid from Step 5). Line $BD$ is the radical axis of the first two circles (their common chord), line $FX$ is the radical axis of the second and third (common chord $FX$ , precisely the shared points), and line $EM$ is the radical axis of the first and third. By the radical axis theorem, since the three circles are pairwise non-concentric, their three pairwise radical axes $BD$ , $FX$ , $EM$ are either all parallel or concurrent; as they are not parallel in this configuration, they are concurrent, which is exactly the desired conclusion.