Is the problem right? Could you please check the original problem? I think there are some typos somewhere. Here is a proof that states : $AC,BD,IJ$ can never concurrent. Let $AC,BD$ intersect at $K$ Let incenter of triangle $ABK, CDK$ be $S,T$. Then it is trivial that $S$ lies on segment $BI$ , $T$ lies on segment $CJ$. But $ST$ passes $K$. So $IJ$ never passes $K$.

teddy8732 wrote: Here is a proof that states : $AC,BD,IJ$ can never concurrent. Wrong.

JG666 wrote: teddy8732 wrote: Here is a proof that states : $AC,BD,IJ$ can never concurrent. Wrong. Why is it wrong? Did I understood problem something wrong? ... JustPostChinaTST wrote: Let $ABCD$ be a convex quadrilateral, the incenters of $\triangle ABD$ and $\triangle ADC$ are $I,J$, respectively. It is known that $AC,BD,IJ$ concurrent at a point $P$. The line perpendicular to $BD$ through $P$ intersects with the inner angle bisector of $\angle ABD$ and the outer angle bisector of $\angle BCD$ at $E,F$, respectively. Show that $PE=PF$. Is $I,J$ are incenters of $\triangle ABD$ and $\triangle ADC$ right? No typos??

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JustPostChinaTST wrote: the incenters of $\triangle ABD$ and $\triangle ADC$ Well, that should be triangle ABC instead of triangle ABD.

This fact kills the problem.

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Just paying attention that ABCD has a Tangential Circle outside（the construction of IMO2008）then it is easy to prove by using polar line and polar point(please forgive my poor English)

This is the solution(without explaination）

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For completeness, here is the full solution: Let $\omega_1$ and $\omega_2$ be the incircles of $\triangle ABC$ and $\triangle ADC$. Claim: $AB+AD=CB+CD$. Proof: Observe that $P$ is the center of negative homothety sending $\omega _1$ to $\omega_2$. Due to symmetry reasons, we may assume that $BC$ cuts $AD$ on the ray $(DA$ beyond $A$, in the point $M$. Consider $\omega$ the excircle of $\triangle DCM$ opposite angle $D$. Then $D$ is the center of positive homothety sending $\omega_2$ to $\omega$. Monge's theorem implies that the center of negative homothety sending $\omega_1$ to $\omega$ lies on line $BD$. However, $BC$ is the internal tangent of $\omega$ and $\omega_1$, thus $B$ is the center we are looking for. This further implies that $AB$ is tangent to $\omega$, making $\omega$ tangent to rays $(DA$ beyond $A$, $(DC$ beyond $C$, $(AB$ beyond $B$ and $(CB$ beyond $B$. Calling these tangency points $U,V,W,T$ respectively, we would have that $$DU=DA+AU=DA+AW=DA+AB+BW$$and similarly $$DV=DC+CV=DC+CT=DC+CB+BT.$$Since $DU=DV$ and $BW=BT$, we get the desired conclusion. Call $\Gamma$ the ellipse with foci in $B$ and $D$ passing through $A$ and $C$. It's well known that the external angle bisectors of angles $BAD$ and $BCD$ are the tangents in $A$ and $C$ to $\Gamma$. Let $d$ be the line through $P$ perpendicular to $BD$, and set $d\cap \Gamma = \{R,S\}$. Since $BD$ is an axis of symmetry in $\Gamma$, $PR=PS$. But now, applying Desargues Involution Theorem to the degenerate quadrilateral $AACC$, conic $\Gamma$ and line $d$, we get that $(P,P)$, $(R,S)$ and $(E,F)$ are involution pairs on $d$, the conclusion following.

Solved with BarisKoyuncu, SerdarBozdag and sevket12. Let $AB \cap CD = X , AD \cap BC = Y , BI \cap DJ = M$. Applying Desargues on triangles $\triangle AIB$ and $\triangle CJD$, we obtain that the points $X , M , G = AI \cap CJ$ are collinear. Now note that since $AI$ is the exterior bisector of $\angle XAC$ and $CJ$ is the bisector of $\angle XCA$, we get that $G$ is the excenter of $\triangle XAC$. So, since $G,X,M$ are collinear, this means $M$ lies on the bisector of $\angle (AB,CD)$, so $d(M,AB)=d(M,CD)$. Since $M$ also lies on $BI$ and $DJ$, we similarly obtain that $d(M,AB)=d(M,BC)=d(M,CD)=d(M,AD)$, implying that the quadrilateral $ABCD$ has an excircle with center $M$. Now we will show that $\angle ADE = \angle CDF$. Note that if we show this, then $A,F$ would be isogonal to $\angle EDC$, and since $E,C$ are obviously isogonal to $\angle EDC$, by isogonality lemma we would obtain $P , M$ are isogonal wrt $\angle EDC$, so this would mean $\angle ADP = \angle JDF$, and so $\angle EDP = \angle FDP$, and from here we would be done. So it indeed suffices to show $\angle ADE = \angle CDF$. Let $H$ and $N$ be the foots from $E$ and $F$ to $DA$ and $DC$ respectively. Since $EHPD$ and $DPFN$ are cyclic, to show $\angle ADE = \angle CDF$, it suffices to show $H,P,N$ are collinear. By Desargues theorem on $\triangle AEH$ and $\triangle CFN$, this collinearity is equivalent to the points $ D , M , O = EH \cap FN$ being collinear. Let $S$ be the intersection of the perpendicular from $D$ to $DA$ with $MA$ and let $T$ be the intersection of the perpendicular from $D$ to $DC$ with $MC$. By Thales, it suffices to show $\frac{MS}{ME} = \frac{MT}{MF}$ which is equivalent to $ST \parallel EF$, or $BD \perp ST$. With this, we get rid of points $E,F$. Let $K,L$ be the foots from $M$ to $CD,AD$ respectively. Let $Z$ be the foot from $M$ to $BD$. We will show that $\triangle KZL \sim \triangle SDT$, and this would imply that $\angle DTS = \angle ZLK = \angle ZDK = 90^\circ - \angle ZDT \implies BD \perp ST$. Also note that $\angle KZL = \angle ADK = 90^\circ - \angle KDS = \angle SDT$, so it suffices to show $\frac{DS}{DT} = \frac{ZK}{ZL}$. Now let $\theta = \angle XAM = \angle MAD$ and $\gamma = \angle YCM = \angle MCD$. We clearly have $DS = DA \cdot \tan{\theta}$ and $DT = DC \cdot \tan{\gamma}$, so $\frac{DS}{DT} = \frac{DA}{DC} \cdot \frac{\tan{\theta}}{\tan{\gamma}}$. By LOS and ratio lemma, we get $\frac{ZK}{ZL} = \frac{\sin{\angle PDC}}{\sin{\angle PDA}} = \frac{PC}{PA} \cdot \frac{DA}{DC}$ , so it suffices to show $\frac{PC}{PA} = \frac{\tan{\theta}}{\tan{\gamma}}$. Note that $\frac{PC}{PA} = \frac{\text{area}(BAD)}{\text{area}(BCD)} = \frac{DA \cdot AB \cdot \sin{2\theta}}{DC \cdot CB \cdot \sin{2\gamma}} = \frac{\tan{\theta}}{\tan{\gamma}} \iff DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$. By law of cosines in $\triangle ABD$ and $\triangle CBD$, we have $DA^2 + AB^2 + 2 \cdot DA \cdot AB \cdot \cos{2\theta} = DC^2 + CB^2 + 2 \cdot DC \cdot CB \cdot \cos{2\gamma}$. Now using the fact that $\cos{2\theta} = 2\cos^2{\theta} - 1$, this gives that $(DA - AB)^2 + 4\cdot DA \cdot AB \cdot \cos^2{\theta} = (DC - CB)^2 + 4\cdot DC \cdot CB \cdot \cos^2{\gamma}$. Since $ABCD$ has an excircle, $DA - AB = DC - CB$ and so we get $DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$, and we are done. $\square$

hakN wrote: Solved with BarisKoyuncu, SerdarBozdag and sevket12. Let $AB \cap CD = X , AD \cap BC = Y , BI \cap DJ = M$. Applying Desargues on triangles $\triangle AIB$ and $\triangle CJD$, we obtain that the points $X , M , G = AI \cap CJ$ are collinear. Now note that since $AI$ is the exterior bisector of $\angle XAC$ and $CJ$ is the bisector of $\angle XCA$, we get that $G$ is the excenter of $\triangle XAC$. So, since $G,X,M$ are collinear, this means $M$ lies on the bisector of $\angle (AB,CD)$, so $d(M,AB)=d(M,CD)$. Since $M$ also lies on $BI$ and $DJ$, we similarly obtain that $d(M,AB)=d(M,BC)=d(M,CD)=d(M,AD)$, implying that the quadrilateral $ABCD$ has an excircle with center $M$. Now we will show that $\angle ADE = \angle CDF$. Note that if we show this, then $A,F$ would be isogonal to $\angle EDC$, and since $E,C$ are obviously isogonal to $\angle EDC$, by isogonality lemma we would obtain $P , M$ are isogonal wrt $\angle EDC$, so this would mean $\angle ADP = \angle JDF$, and so $\angle EDP = \angle FDP$, and from here we would be done. So it indeed suffices to show $\angle ADE = \angle CDF$. Let $H$ and $N$ be the foots from $E$ and $F$ to $DA$ and $DC$ respectively. Since $EHPD$ and $DPFN$ are cyclic, to show $\angle ADE = \angle CDF$, it suffices to show $H,P,N$ are collinear. By Desargues theorem on $\triangle AEH$ and $\triangle CFN$, this collinearity is equivalent to the points $ D , M , O = EH \cap FN$ being collinear. Let $S$ be the intersection of the perpendicular from $D$ to $DA$ with $MA$ and let $T$ be the intersection of the perpendicular from $D$ to $DC$ with $MC$. By Thales, it suffices to show $\frac{MS}{ME} = \frac{MT}{MF}$ which is equivalent to $ST \parallel EF$, or $BD \perp ST$. With this, we get rid of points $E,F$. Let $K,L$ be the foots from $M$ to $CD,AD$ respectively. Let $Z$ be the foot from $M$ to $BD$. We will show that $\triangle KZL \sim \triangle SDT$, and this would imply that $\angle DTS = \angle ZLK = \angle ZDK = 90^\circ - \angle ZDT \implies BD \perp ST$. Also note that $\angle KZL = \angle ADK = 90^\circ - \angle KDS = \angle SDT$, so it suffices to show $\frac{DS}{DT} = \frac{ZK}{ZL}$. Now let $\theta = \angle XAM = \angle MAD$ and $\gamma = \angle YCM = \angle MCD$. We clearly have $DS = DA \cdot \tan{\theta}$ and $DT = DC \cdot \tan{\gamma}$, so $\frac{DS}{DT} = \frac{DA}{DC} \cdot \frac{\tan{\theta}}{\tan{\gamma}}$. By LOS and ratio lemma, we get $\frac{ZK}{ZL} = \frac{\sin{\angle PDC}}{\sin{\angle PDA}} = \frac{PC}{PA} \cdot \frac{DA}{DC}$ , so it suffices to show $\frac{PC}{PA} = \frac{\tan{\theta}}{\tan{\gamma}}$. Note that $\frac{PC}{PA} = \frac{\text{area}(BAD)}{\text{area}(BCD)} = \frac{DA \cdot AB \cdot \sin{2\theta}}{DC \cdot CB \cdot \sin{2\gamma}} = \frac{\tan{\theta}}{\tan{\gamma}} \iff DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$. By law of cosines in $\triangle ABD$ and $\triangle CBD$, we have $DA^2 + AB^2 + 2 \cdot DA \cdot AB \cdot \cos{2\theta} = DC^2 + CB^2 + 2 \cdot DC \cdot CB \cdot \cos{2\gamma}$. Now using the fact that $\cos{2\theta} = 2\cos^2{\theta} - 1$, this gives that $(DA - AB)^2 + 4\cdot DA \cdot AB \cdot \cos^2{\theta} = (DC - CB)^2 + 4\cdot DC \cdot CB \cdot \cos^2{\gamma}$. Since $ABCD$ has an excircle, $DA - AB = DC - CB$ and so we get $DA \cdot AB \cdot \cos^2{\theta} = DC \cdot CB \cdot \cos^2{\gamma}$, and we are done. $\square$ $F$

Nice problem! Claim: The two internal bisectors of $\angle ABC$, $\angle ADC$ and two external of $\angle BAD$, $\angle BCD$ are concurrent, i.e. there's an excircle. Proof. Let the first two intersect at $O$ and draw $\omega$ centered at $O$ tangent to $BA$ and $BC$. By Monge d'Alembert's theorem, the insimilicenter of $\omega$ and the incircle of $\triangle ADC$ lies on $BD$. But since $D\in JO$ it follows that the insimilicenter is $D$. So $DA$ and $DC$ are tangent to $\omega$ and the claim follows. Now, it's well known that $BC+CD=BA+AD$ which means that $A$ and $C$ lie on an ellipse $\mathcal{E}$ with foci $B$, $D$. Then \[(E,F;P,\infty_{\perp BD})\stackrel{O}{=}(A,C;P,OX\cap AC)\]where $X$ is the projection of $O$ onto $BD$. It remains to show that $PX$ is the bisector of $\angle AXC$ and we'd be done by the perpendicularity-angle bisector lemma. To this end, note $AO$ and $CO$ are tangents to $\mathcal{E}$ and if $A'$ and $C'$ are reflections of $A$ and $C$ over the major axis, Brokard's theorem on $AA'C'C$ finishes.

Claim: We have $AB+AD=CB+CD$. Proof. By Pitot's theorem twice, proving that there is a circle tangent to rays $\overrightarrow{AB}$, $\overrightarrow{CB}$, $\overrightarrow{DA}$, and $\overrightarrow{DC}$ is sufficient. Let $Q = \overline{BC} \cap \overline{AD}$, and denote $\omega$ by the $D$-excircle of $\triangle DQM$ and $O$ by the center of $\omega$. I contend that $\omega$ is the desired circle. Let $\omega_B$ be the incircle of $\triangle ABC$ and $\omega_D$ be the incircle of $\triangle ADC$. By Monge's theorem, the insimilicenter $\textbf{I}(\omega_B, \omega)$ of $\omega_B$ and $\omega$ lies on $\overline{BD}$, since $D$ is the insimilicenter of $\omega_D$ and $\omega$. Since $\textbf{I}(\omega_B, \omega)$ also lies on $\overline{OJ}$, it must be $D$. Thus the desired tangencies follow. Note that by the claim, there exists an ellipse $\epsilon$ with foci at $B$ and $D$ passing through $A$ and $C$. Define $\ell$ as the line through $P$ perpendicular to $\overline{BD}$, and intersect $\epsilon$ at points $X$ and $Y$. Clearly we have $PX=PY$. Applying DIT on $\epsilon$ with respect to $\ell$, there exists an involution swapping $(P, P)$, $(E, F)$, and $(X, Y)$, but this means that $PE=PF$, as desired.

Claim: $ABCD$ is extangential. Proof. By Desargues on $AIB$ and $CJD$ it follows that $H = \overline{AI} \cap \overline{CJ}$, $O = \overline{IB} \cap \overline{JD}$, and $G = \overline{AB} \cap \overline{CD}$ are collinear. It follows that $H$ is the $C$-excenter of $ACG$, so $HG$ is the external bisector of $AGD$, so $O$ is the desired excenter. $\blacksquare$ Then, the external angle bisectors of $BAD$ and $BCD$ are $AO$ and $CO$ respectively. Let $Q$ be the polar of $BD$ wrt to the excircle. Then it remains to show that $(AC;PQ) = -1$. Polar reciprocating the diagram gives the result with taking $P$ as the intersection of the diagonals.

Why are so many CTST geos projective spam? The following is the key claim: Key Claim: $ABCD$ is extangential quadrilateral.

proof

Now let $\mathcal{E}$ be an ellipse focussed at $B$ and $D$ passing through $A$ and $C$.Let $\mathcal{l}$ denote the line through $P$ perpendicular to $BD$ Extend line $BD$ to $\mathcal{E}$ at $U,V$. Apply DIT on $\mathcal{E}$ WRT $\mathcal{l}$ and realise you are done.

Remark from 数之谜: After $|AB|+|AD|=|CB|+|CD|,$ $A,C$ moves on an ellipse $\Gamma$ with focus $B,D.$ Since $AE$ is the outer angle bisector of $\angle BAD,$ $AE$ is tangent line of $\Gamma,$ similarly $CF$ is also tangent line of $\Gamma.$ Butterfly Theorem gives $|PE|=|PF|.\Box$

Let $X,Y=AB\cap CD,AD\cap BC$. Say $X,Y$ lie on same side of $AC$ as $B$. Construct circle $\Omega$ with center $Z$ tangent to extensions of rays $AB,CB,CX$. By insimilicenter monge on $\Omega,(I),(J)$ we get $\Omega$ also tangent to $AY$. Now dual brocard's gives that dual of $BD$ wrt $\Omega$ is $AC\cap XY=Q$. Thus $-1=(A,C;P,Q)\stackrel Z=(E,F;P,\infty)$ finishes.