Problem 2

Let

BC

be a diameter of circle

\omega

with center

O

. Let

A

be a point of

\omega

such that

0^\circ<\angle AOB<120^\circ

. Let

D

be the midpoint of arc

AB

not containing

C

. The line

\ell

passes through

O

and is parallel to line

AD

. Line

\ell

intersects line

AC

J

. The perpendicular bisector of segment

OA

intersects circle

\omega

E

and

F

. Prove that

J

is the incenter of triangle

CEF

Step 1 of 3: Identify the rhombus

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In plain words

The perpendicular-bisector chord of $OA$ creates two equilateral triangles.

AE=AF=AO=EO=OF=s

Detailed analysis

Let $M$ be the midpoint of $EF$ . Since $EF$ is the perpendicular bisector of $OA$ , $M$ is also the midpoint of $OA$ . For either $X=E$ or $X=F$ , we have $XO=XA$ because $X$ lies on the perpendicular bisector, and $XO=AO$ because $X$ lies on $\omega$ . Thus $AEO$ and $AFO$ are equilateral triangles. Consequently $AEOF$ is a rhombus, and with $s=AO$ the five displayed segments have the common length $AE=AF=AO=EO=OF=s$ .