Problem 6

Let

I

be the incenter of acute triangle

ABC

with

AB\ne AC

. The incircle

\omega

ABC

is tangent to

BC

CA

, and

AB

D

E

, and

F

, respectively. The line through

D

perpendicular to

EF

meets

\omega

again at

R

(other than

D

). Line

AR

meets

\omega

again at

P

(other than

R

). The circumcircles of triangles

PCE

and

PBF

meet again at

Q

(other than

P

). Prove that lines

DI

and

PQ

meet on the line through

A

perpendicular to

AI

Step 1 of 6: Name the target point

In plain words

The claimed point is the intersection of two fixed lines.

T=DI\cap\{X:AX\perp AI\}

Detailed analysis

Let $T$ be the intersection of $DI$ with the line through $A$ perpendicular to $AI$ . The statement is equivalent to proving that $T$ lies on $PQ$ .