$\delta$ is the circumscribed circle on a cyclic quadrilateral ABCD. The centre of the inscribed circle of triangle ABC is P, and that of triangle ABD is Q. Let E denote the midpoint of arc BC, and let F denote the midpoint of arc DA of the circle $\delta$.

Prove that the line segment PQ is parallel to EF.


First note the following well known result:

Lemma: Let $ABC$ be a triangle with incenter $I$, and excenter $I_A$ opposite $A$. The line $AII_A$ intersects the circumcircle of triangle $ABC$ at another point $D$ (other than $A$). Then $D$ is the midpoint of arc $BC$ (not containing $A$ of the circumcircle and $DI=DB=DC=DI_A$.

I have provided the proof just in case you do not know this result. Proof: Note that by definition both $I, I_A$ lie on internal angle bisector of $\angle(BAC)$, so $A, I, I_A$ are indeed collinear.

$$\angle ICD =\angle ICB +\angle BCD =\angle ICA +\angle BAD =\angle ICA +\angle CAI =\angle DIC$$

Note that $CI \perp CI_A$, so

$$\angle DCI_A=90^{\circ}-\angle ICD=90^{\circ}-\frac{180^{\circ}-\angle IDC}{2}=\frac{\angle IDC}{2}$$

Thus $DI=DC=DI_A$. Similarly $DI=DB=DI_A$, so we also have $DB=DC$ and $D$ is the midpoint of arc $BC$, so we are done.

Let's continue. Let $G$ be the midpoint of arc $AB$ (not containing $C, D$) By the above lemma, $A, P, E$ are collinear and $PE=PB=PC$, $B, Q, F$ are collinear and $FQ=FA=FC$, $C, P, G$ are collinear and $GP=GA=GB$, $D, Q, G$ are collinear and $GQ=GA=GB$. Note that $GP=GQ$.

For convenience, let us denote the angle subtended by an arc $XY$ as $\angle\widehat{XY}$. We now angle chase.

\begin{align} \angle FQA=\frac{180^{\circ}-\angle \widehat{AGB}}{2} & ; &\angle AQG=\frac{180^{\circ}-\angle \widehat{DFA}}{2} \end{align}


$$\angle GQB=180^{\circ}-\angle FQA-\angle AQG=\frac{\angle \widehat{AGB}+\angle \widehat{DFA}}{2}=\frac{\angle \widehat{DAB}}{2}$$

$$\angle GQP=\frac{180^{\circ}-\angle \widehat{CD}}{2}=\frac{\angle \widehat{DABC}}{2}$$

$$\angle BQP=\angle GQP-\angle GQB=\frac{\angle \widehat{DABC}}{2}-\frac{\angle \widehat{DAB}}{2}=\frac{\angle \widehat{BEC}}{2}=\angle \widehat{BE}=\angle BFE$$

Thus $PQ \parallel EF$ and we are done.


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