Problem 4

Let

ABCD

be a quadrilateral with an incircle

\omega

of centre

I

. The diagonals

AC

and

BD

intersect at

E

. Let

J

be the incentre of triangle

ABD

. The extension of the ray

EJ

intersects

\omega

P

. Prove that

PI\perp BD

Step 5 of 6: Chase the cross-ratio to force collinearity

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\text{pole--polar and inversion calculation} \implies J(IV,P'S)=A(SE,VJ)=J(SE,VA)=J(AV,ES)=J(IV,ES) \implies E,J,P'\text{ collinear}

Detailed analysis

The pole–polar and inversion calculation applied to the harmonic range in Step 4 gives the chain $J(IV,P'S)=A(SE,VJ)=J(SE,VA)=J(AV,ES)=J(IV,ES)$ . Here $A,Q,S$ are collinear by Step 2 and $BD$ is the polar of $L$ by Step 3. The final equality compares two cross-ratios in the same pencil at $J$ and forces the lines $JP'$ and $JE$ to coincide; hence $E,J,P'$ are collinear.