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 2 of 3: Locate

J

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In plain words

Parallel lines turn the construction into a parallelogram.

AJ=AO=s

Detailed analysis

Put $\alpha=\angle BOD=\angle DOA$ , so $\angle BOA=2\alpha$ . Since $BC$ is a diameter, triangle $AOC$ is isosceles and $\angle OAC=\angle ACO=\alpha$ . Because $J$ lies on $AC$ , $\angle OAJ=\angle OAC=\alpha=\angle AOD$ . Together with $OJ\parallel AD$ , this gives the parallelogram $DAJO$ : indeed $OJ\parallel AD$ and $AJ\parallel DO$ . Hence $M$ , the midpoint of $AO$ , is also the midpoint of $DJ$ , so $J$ is the reflection of $D$ in $M$ . To obtain the length, triangle $AOD$ is isosceles with vertex angle $\alpha$ , so $\angle OAD=90^\circ-\alpha/2$ . Since $OJ\parallel AD$ , $\angle AOJ=90^\circ-\alpha/2$ ; in triangle $AOJ$ the remaining angle $\angle AJO$ is also $90^\circ-\alpha/2$ . Therefore $\angle AJO=\angle AOJ$ , and $AJ=AO=s$ .