Since $ BI, $ $ DJ $ passes through the midpoint of arc $ ADC, $ arc $ CBA $ in $ \odot (ABCD) $, respectively, so from Reim's theorem we get $ IJ $ $ \perp $ $ AC $ $ \Longrightarrow $ the tangency point of $ \odot (I) $ with $ AC $ coincide with the tangency point of $ \odot (J) $ with $ AC $, hence if $ H $ is the intersection of $ AC $ and $ IJ $, then $\tfrac{1}{2} (AB+AC-BC) $ $ = $ $ AH $ $ = $ $ \tfrac{1}{2} $ $ (AD+AC-CD) $ $ \Longrightarrow $ $ AB $ $ + $ $ CD $ $ = $ $ AD $ $ + $ $ BC $. i.e. there is a circle with center $ T $ $ \equiv $ $ BI $ $ \cap $ $ DJ $ and tangent to the sides of $ ABCD $ Since $ \angle CTA $ $ + $ $ \angle CXA $ $ = $ $ \tfrac{1}{2} \angle A $ $ + $ $ \tfrac{1}{2} \angle C $ $ + $ $ \angle B $ $ + $ $ 90^{\circ} $ $ = $ $ 180^{\circ} $ $ + $ $ \angle B $, so notice $ T $ and $ X $ lie on the bisector of $ \angle B $ we get $ T, $ $ X $ are isogonal conjugate WRT $ \triangle ABC $ $ \Longrightarrow $ $ \angle XAC $ $ = $ $ \angle TAB $. Similarly, we can prove $ \angle YAC $ $ = $ $ \angle TAD $, so $ \angle XAC $ $ = $ $ \angle YAC $ $ \Longrightarrow $ $ X $ and $ Y $ are symmetry WRT $ AC $.

Another way to deduce that $ABCD$ is tangential is as follows: Let $K$ be the incenter of $\triangle ABD$ and let $M$ be the midpoint of arc $\widehat{AB}$ on $\odot(ABCD).$ By the Incenter/Excenter Lemma, $M$ is the circumcenter of both $\triangle AIB$ and $\triangle AKB.$ Therefore, $AKIB$ is cyclic. Similarly $AKJD$ is cyclic. Then $T \equiv AK \cap BI \cap DJ$ is the radical center of $\odot(AIKB), \odot(AJKD), \odot(BIJD).$ As $T$ lies on the bisectors of $\angle A, \angle B, \angle D$, it follows that $ABCD$ has an incircle centered at $T.$

TelvCohl wrote: Since $ BI, $ $ DJ $ passes through the midpoint of arc $ ADC, $ arc $ CBA $ in $ \odot (ABCD) $, respectively, so from Reim's theorem we get $ IJ $ $ \perp $ $ AC $ $ \Longrightarrow $ the tangency point of $ \odot (I) $ with $ AC $ coincide with the tangency point of $ \odot (J) $ with $ AC $, hence if $ H $ is the intersection of $ AC $ and $ IJ $, then $\tfrac{1}{2} (AB+AC-BC) $ $ = $ $ AH $ $ = $ $ \tfrac{1}{2} $ $ (AD+AC-CD) $ $ \Longrightarrow $ $ AB $ $ + $ $ CD $ $ = $ $ AD $ $ + $ $ BC $. i.e. there is a circle with center $ T $ $ \equiv $ $ BI $ $ \cap $ $ DJ $ and tangent to the sides of $ ABCD $ Since $ \angle CTA $ $ + $ $ \angle CXA $ $ = $ $ \tfrac{1}{2} \angle A $ $ + $ $ \tfrac{1}{2} \angle C $ $ + $ $ \angle B $ $ + $ $ 90^{\circ} $ $ = $ $ 180^{\circ} $ $ + $ $ \angle B $, so notice $ T $ and $ X $ lie on the bisector of $ \angle B $ we get $ T, $ $ X $ are isogonal conjugate WRT $ \triangle ABC $ $ \Longrightarrow $ $ \angle XAC $ $ = $ $ \angle TAB $. Similarly, we can prove $ \angle YAC $ $ = $ $ \angle TAD $, so $ \angle XAC $ $ = $ $ \angle YAC $ $ \Longrightarrow $ $ X $ and $ Y $ are symmetry WRT $ AC $. What is Reim's theorem? I google it but probably it was not same with your used one.

There is only one Reim's theorem as far as I'm concerned (a simple google search produces this and this), and it is the same as Telv is using here. Considering the circles $(BIJD),(ABCD)$, which intersect at $B,D$, we know that if $BI,DJ$ intersect $(ABCD)$ again at $M,N$, which are the midpoints of arcs $\widehat{ADC},\widehat{CBA}$ of $(ABCD$, then $IJ\parallel MN$. But $MN\perp AC$, so $IJ\perp AC$ too.

Solution that avoids proving $ABCD$ has an incircle" Let $N,M$ be the midpoints of arcs $ABC,ADC$. By Reim's theorem(or angle chasing), $I,J,D,B$ are concyclic $\implies (IJ||NM)\perp AC$. Thus $IM,IJ$ are isogonal wrt $\angle AIC$ because $M$ is the circumcenter of $\triangle IAC$. In fact, we can verify that $J,X$ are isogonal conjugates wrt $\triangle IAC$ using angle relations and uniqueness. This implies $\angle IAX=\angle CAJ$. Similarly $I,Y$ are isogonal conjugates wrt $\triangle JAC$ $\implies \angle IAC=\angle JAY$. It is now apparent that $X,Y$ are symmetrical over $AC$ by noting \[\angle XAC=\angle XAI+\angle IAC=\angle CAJ+\angle JAY=\angle CAY.\]

buzzychaoz wrote: In cyclic quadrilateral $ABCD$, $AB>BC$, $AD>DC$, $I,J$ are the incenters of $\triangle ABC$,$\triangle ADC$ respectively. The circle with diameter $AC$ meets segment $IB$ at $X$, and the extension of $JD$ at $Y$. Prove that if the four points $B,I,J,D$ are concyclic, then $X,Y$ are the reflections of each other across $AC$. Nice problem! Let $T=BI \cap DJ$. Then $TB, TD$ bisect arcs made by chord $AC$; hence by Reim's Theorem, $BIJD$ cyclic yields $IJ \perp AC$. Let $S=IJ \cap AC$. Thus, $AB-BC=AS-SC=AD-DC$ yielding that $ABCD$ is bi-centric. Now we observe that $\angle AXC+\angle ATC=180^{\circ}+\angle ABC$ and $T$ lies on $XB$; hence $X,T$ are isogonal conjugates in $\triangle ABC$. Consequently, $\angle XAC=\angle TAB$. Likewise $\angle YAC=\angle TAD$ and $T$ lies on the bisector of angle $DAB$; thus proving the result. $\blacksquare$

Nice problem but easy !

buzzychaoz wrote: In cyclic quadrilateral $ABCD$, $AB>BC$, $AD>DC$, $I,J$ are the incenters of $\triangle ABC$,$\triangle ADC$ respectively. The circle with diameter $AC$ meets segment $IB$ at $X$, and the extension of $JD$ at $Y$. Prove that if the four points $B,I,J,D$ are concyclic, then $X,Y$ are the reflections of each other across $AC$. Well this one took me lotta of timeeee, like >3 hrs so like i dieeeeeeeee. The hardest part of the problem is to know what u are doing, ok so let $E$ the incenter of $\triangle BCD$ and let $G$ the incenter of $\triangle ABD$ then clearly $BIEC, CEJD, JDGA, AGIB$ are cyclic and by radax using that $BIJD$ is cyclic we get that $ABCD$ is bicentric, let $Z$ be the incenter of $ABCD$ then since $\angle BAZ+\angle BCZ=90$ we get $\angle AZC=90+\angle ABC$ and by more angle chasing u get $\angle XAI+\angle XCI=\frac{\angle ABC}{2}=\angle IAZ+\angle ICZ$ and with this u get $X,Z$ isogonal conjugates w.r.t. $\triangle ABC$ and in a similar way u can get $Z,Y$ isogonal conjugates w.r.t. $\triangle ADC$ now using angle bisector theorem $\frac{AX}{XC}=\frac{AZ}{ZC}=\frac{AY}{YC}$ and since $\angle AXC=90=\angle AYC$ u get $\triangle AXC \sim \triangle AYC$ but since they share $AC$ as common side, this similarity turns out to be a congruency, so u get $AX=AY$ and $CX=CY$ which means $X,Y$ symetric w.r.t. $AC$ as desired, thus we are done P.S. 800th post!? when did that happen...

Let $M$ and $N$ be the arc midpoints of $AC$ opposite $B$ and $D$ respectively. Then by Reim's theorem, we have $IJ\parallel MN$ which is perpendicular to $AC$. It is well-known that this implies that $ABCD$ has an incircle - let its incentre be $T$. Since $\angle MAT+\angle MCT=\frac{1}{2}(\angle MAB+\angle MCB)=90^{\circ}=\angle AXC$, the inversion centred at $M$ with radius $MA$ swaps $T$ and $X$. Let $MB$ intersect $AC$ at $H$, so $H$ swaps with $B$ under the inversion. Then, we have $$\angle TAB=\angle MAB-\angle MAT=\angle MHA-\angle MXA=\angle HAX=\angle CAX$$Similarly, using the inversion centred at $N$ with radius $NA$, we have $\angle TAD=\angle CAY$. Therefore, $\angle CAX=\angle TAB=\angle TAD=\angle CAY$ so $X$ and $Y$ are reflections over $AC$ since they lie on the circle with diameter $AC$.