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 6 of 7: Reflection across

EM

gives the concurrency

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In plain words

Once $EX\parallel FM$ is known, equal legs are all that is needed for an isosceles trapezoid, and those legs come from another instance of the repeated angle $\alpha$ .

B,X\text{ and }D,E\text{ are symmetric about }EM\quad\Rightarrow\quad BD\text{ and }FX\text{ meet on }EM

Detailed analysis

The line $EM$ bisects $\angle FMD$ , so it is the symmetry axis of the isosceles trapezoid $BFDX$ . Thus $B,X$ are symmetric about $EM$ , and $D,E$ are symmetric about $EM$ . The reflection of line $BD$ is therefore line $FX$ ; these two lines are concurrent at their intersection on the axis $EM$ . The axis is $EM$ .