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Let $ABC$ be a right-angled triangle with $\widehat{B}=90^\circ$ and let $BD$ be the altitude from $B$ on to $AC$. Draw $DE\perp AB$ and $DF\perp BC$. Let $P, Q, R$ and $S$ be respectively the incentres of triangle $DFC$, $DBF$, $DEB$ and $DAE$. Suppose $S, R, Q$ are collinear. Prove that $P, Q, R, D$ lie on a circle.

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  • $\begingroup$ Please improve formatting and add some context/attempts. $\endgroup$ – Jack D'Aurizio Nov 12 '17 at 13:12
  • $\begingroup$ $ADB$ and $BDC$ map to each other through a spiral similarity centered at $D$. This allows to perform some angle chasing. $\endgroup$ – Jack D'Aurizio Nov 12 '17 at 13:46
  • $\begingroup$ In particular, in the given configuration $\widehat{SDQ},\widehat{RDQ}$ and $\widehat{PQR}$ all are right angles. $\endgroup$ – Jack D'Aurizio Nov 12 '17 at 13:51
  • $\begingroup$ I tried to use coordinate geometry by assuming angle A as and hyp to be fixed which are general property of any triangle. Hyp is divided in k:1 ratio, finding all points and incenters by incenter formula we get all points. Then prove what is required is possible but Is too lengthy. Want to use geometry please help $\endgroup$ – Shubham Patel Nov 12 '17 at 15:11
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As pointed out in the comments, a spiral similarity around $D$ maps $ADB$ into $BDC$, hence $\widehat{RDP}$ is surely a right angle. In order to prove the statement, it is enough to show that $\widehat{PQR}=\widehat{PQS}$ is a right angle too. On the other hand the previous spiral similarity brings $S$ into $Q$ and $R$ into $P$ by rotating around $D$ by $90^\circ$ counterclockwise, then applying a suitable dilation. It follows that the $SR$ line is orthogonal to the $QP$ line and we are done: $PR$ is a diameter of the circumcircle of $PQRD$.

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