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 2 of 6: Locate

Q

via a homothety at

A

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A,Q,S\text{ collinear, where }S=(J)\cap BD

Detailed analysis

Let $S$ be the tangency point of the incircle $(J)$ of triangle $ABD$ with $BD$ . Both circles are tangent to the rays $AB,AD$ , so a homothety centred at $A$ maps $\omega$ to $(J)$ and maps $Q$ to $S$ . Hence $A,Q,S$ are collinear.