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 1 of 6: Reformulate the goal

Previous step Next step

P'Q\text{ is the diameter of }\omega\text{ with }P'Q\perp BD,\quad AP'<AQ;\quad P=P'\Longleftrightarrow E,J,P'\text{ collinear}

Detailed analysis

Choose the diameter $P'Q$ of $\omega$ perpendicular to $BD$ , labelling its endpoints so that $AP'<AQ$ . Since $PI\perp BD$ exactly when $P$ is an endpoint of this diameter, it is enough to prove that the second intersection $P$ of ray $EJ$ with $\omega$ is $P'$ , equivalently that $E,J,P'$ are collinear.