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 4 of 6: Use the antipode and parallelogram

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In plain words

The antipode of the contact point reveals a hidden line through $P$ .

P,A,G\text{ are collinear},\qquad M=\operatorname{center}(DCAB)

Detailed analysis

Let $G$ be the antipode of $D$ on $\omega$ . The harmonic quadrilateral determined by $E,F,P,R$ and projection through $G$ gives $P,A,G$ collinear. Let $M$ be the center of parallelogram $DCAB$ ; then $M$ is the midpoint of $BC$ and $MI$ is parallel to $AG$ . The circle with diameter $DA$ passes through $P$ and $T$ , because $\angle DPA=90^\circ$ and $DT\perp AT$ .