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 2 of 6: Invert in the incircle

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In plain words

The contact triangle makes the inverse configuration linear.

A'=\operatorname{mid}(EF),\quad B'=\operatorname{mid}(FD),\quad C'=\operatorname{mid}(DE)

Detailed analysis

Invert about $\omega$ with center $I$ and radius equal to its inradius. Since $AE$ and $AF$ are tangents, the inverse of $A$ is the midpoint $A'$ of $EF$ ; similarly the inverses of $B,C$ are the midpoints $B'$ of $FD$ and $C'$ of $DE$ . Points on $\omega$ , including $D,E,F,P,R$ , remain fixed. The standard inversion reduction (or direct angle chase) says that the original assertion is equivalent to showing that the inverse image $Q'$ lies on the circle $(PIT)$ in the inverted figure.