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 6 of 6: Conclude

PI \perp BD

Previous step Next step

E,J,P'\text{ collinear} \implies P=P' \implies PI=P'I\parallel\text{(diameter }P'Q\perp BD) \implies PI\perp BD

Detailed analysis

By Step 5, $E$ , $J$ and $P'$ are collinear, so the ray $EJ$ extended meets $\omega$ exactly at $P'$ ; that is, $P=P'$ . Since $P'$ was chosen so that $P'Q$ is a diameter of $\omega$ perpendicular to $BD$ , the segment $PI$ lies along this diameter, and therefore $PI\perp BD$ .