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 4 of 7: The parallelogram puts

X

on a circle

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In plain words

The equal angle $\alpha$ at $D$ and at $B$ places $E$ symmetrically on the circle through $A,B,M,D$ .

MX=EA=MF,\quad B,C,D,F,X\text{ concyclic}

Detailed analysis

The parallelogram gives $MX=EA=MF$ . Since $M$ is the midpoint of $CF$ , we have $MC=MF=DE$ ; also $DE\parallel MC$ . Thus $MCDE$ is a parallelogram, and $B,C,D,F,X$ lie on one circle (equivalently, the equal chords obtained from $MX=MF$ give the same cyclic quadrilateral).