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 3 of 6: Pascal's theorem places

L

AC

and identifies its polar

Previous step Next step

W,X,Y,Z=\omega\cap AB,BC,CD,DA;\quad L:=XY\cap ZW \implies L\in AC,\quad BD=\text{polar}_\omega(L)

Detailed analysis

Let $W,X,Y,Z$ be the points where $\omega$ touches $AB$ , $BC$ , $CD$ , $DA$ respectively, and set $L:=XY\cap ZW$ . Applying Pascal's theorem to the (degenerate) hexagon inscribed in $\omega$ through these contact points, taken with the appropriate repetitions along the tangent lines, shows that $L\in AC$ ; combined with the standard pole-polar correspondence for the tangential quadrilateral, this also shows that $BD=\text{polar}_\omega(L)$ .