Suppose that P is closer to D. Let I1, I2, J1, J2 be the incircle of ACD, incircle of ACB, D-excircle of ACD and B-excircle of ACB, respectively, by 2017G7, J1, J2 and AC are concurrent at U, and I1, I2 and AC are also concurrent. By homothety, D, P, U are colinear, so done.

Let $\omega$ touches $AB, BC, CD, DA$ at $E, F, G, H$ respectively. Clearly $EH, FG, BD, PQ$ are concurrent at pole $T$ of $AC$. Let $BX, DX$ intersect $\omega$ at $U, V$ respectively. Notice that $$(EF;PU) = -1 = (GH;QV)$$so projecting through $T$, points $U, V, T$ are colinear. Hence by Brokard's theorem we can conclude that $X=PU\cap QV$ lies on polar of $T$ w.r.t. $\omega$.

Let $M,N,R,S$ be the tangency points on $AB,BC,CD,DA,$ respectively. Well-known that $MS, RN, PQ$ concur (at $E, AC$ is polar of $E$, $B,D,E$ are also collinear because $BD$ is a radical axis of the circles $MIS,RIN$...). Also $AC,MP,QS$ concur and also $AC,MQ,PS$ because of well-known properties of the cyclic quadrilateral $MQPS$. The end follows by Desargues on $\Delta BQM,\Delta DPS$ and on $\Delta BPM,\Delta DQS$.

Notice that $P$ is the top-point of $\odot(I)$ and $\odot(I)$ is similar to incircle of $\triangle ABC$ under dilation at $B$. Consequently, line $\overline{BP}$ passes through the touch-point of $B$-excircle on $\overline{AC}$. Likewise line $\overline{BQ}$ passes through $B$-intouch point on $\overline{AC}$. Finally, remark that $AB-BC=AD-DC$; concluding the proof.

MarkBcc168 wrote: Let $\omega$ touches $AB, BC, CD, DA$ at $E, F, G, H$ respectively. Clearly $EH, FG, BD, PQ$ are concurrent at pole $T$ of $AC$. Let $BX, DX$ intersect $\omega$ at $U, V$ respectively. Notice that $$(EF;PU) = -1 = (GH;QV)$$so projecting through $T$, points $U, V, T$ are colinear. Hence by Brokard's theorem we can conclude that $X=PU\cap QV$ lies on polar of $T$ w.r.t. $\omega$. Can you use perspectivity like that? I think that you have to have point $T$ on the circle ?

FISHMJ25 wrote: MarkBcc168 wrote: Let $\omega$ touches $AB, BC, CD, DA$ at $E, F, G, H$ respectively. Clearly $EH, FG, BD, PQ$ are concurrent at pole $T$ of $AC$. Let $BX, DX$ intersect $\omega$ at $U, V$ respectively. Notice that $$(EF;PU) = -1 = (GH;QV)$$so projecting through $T$, points $U, V, T$ are colinear. Hence by Brokard's theorem we can conclude that $X=PU\cap QV$ lies on polar of $T$ w.r.t. $\omega$. Can you use perspectivity like that? I think that you have to have point $T$ on the circle ? Yes you can, and no, $T$ does not have to be on circle if you project from circle to the same circle

rmtf1111 wrote: FISHMJ25 wrote: MarkBcc168 wrote: Let $\omega$ touches $AB, BC, CD, DA$ at $E, F, G, H$ respectively. Clearly $EH, FG, BD, PQ$ are concurrent at pole $T$ of $AC$. Let $BX, DX$ intersect $\omega$ at $U, V$ respectively. Notice that $$(EF;PU) = -1 = (GH;QV)$$so projecting through $T$, points $U, V, T$ are colinear. Hence by Brokard's theorem we can conclude that $X=PU\cap QV$ lies on polar of $T$ w.r.t. $\omega$. Can you use perspectivity like that? I think that you have to have point $T$ on the circle ? Yes you can, and no, $T$ does not have to be on circle if you project from circle to the same circle My mistake , you just fix circle with inversion and thats it.

Solved with Jeffrey Chen, Max Lu, and Raymond Feng. Without loss of generality \(P\) is on the same side of \(\overline{AC}\) as \(B\), and \(Q\) is one the same side of \(\overline{AC}\) as \(D\). Let the incircles of \(\triangle ABC\) and \(\triangle ADC\) touch \(\overline{AC}\) at a common point \(T\) by Pitot. Then since \(\overline{PP}\parallel\overline{QQ}\parallel\overline{AC}\), by homothety we have \(B\), \(T\), \(Q\) are collinear Analogously \(D\), \(T\), \(P\) are collinear, so \(T=\overline{BQ}\cap\overline{DP}\in\overline{AC}\). Analogously \(\overline{BP}\cap\overline{DQ}\in\overline{AC}\), as desired.

Let $AB,BC,CD,DA$ touch to $\omega$ at $X,Y,Z,T$. Let $K=TX\cap YZ, R=AC\cap PQ, F=QC\cap AD, G=PC\cap AB$. At first not that all poles and polars are taken wrt $\omega$.It's well-known that $K\in BD$. $RITX,RIYZ, TXYZ$ are cyclic, so from Radical Axis theroem $K\in PQ$. $K$ lies on polar of $A$ and $C$, so from La-Hire we get $AC$ is polar of $K$. Applying Brokard's theorem on $PQXT$, gives that $TQ, XP,AC$ are concurrent. Since $TQ\cap XP, TF\cap XG, FQ\cap GP$ are collinear, we get triangles $TFQ$ and $XGP$ are perspective. So $TX, FG, QP$ are concurrent, i.e $K,F,G$ are collinear. So $DBA$ and $QPC$ are perspective $\implies DQ\cap BP\in AC \implies X\in AC$. Since $AC$ is polar of $K$, $(K,R;P,Q)=-1$. Projecting it to $AC$ through $D$ and $B$ gives that $(BD\cap AC,R;DP\cap AC,X)=-1$ and $(BD\cap AC,R; BQ\cap AC,X)=-1$, respectively. So $BQ\cap AC\equiv DP\cap BC \implies Y\in AC$. So we are done.

It is easy to see, using the length condition, that the incircles in $\triangle ACD$ and $\triangle ABC$ touch at a point $X$ on $AC$. Notice that the tangent at $Q$ to $\omega$ is parallel to $AC$, so the homothety at $D$ taking the incircle of $\triangle ACD$ to $\omega$ takes $X$ to $Q$, so $X$ lies $DQ$. Similarly $X$ lies on $BP$. The same approach can be applied for $Y$ - the common point of the excircles of the two triangles.

Here is a simple solution using Pascal and pole-polar duality.

The condition is equivalent to proving that the pole of $AC$ lies on the polar of $X$. It is well know that the pole of $AC$ is $M$. Now notice that $QQ$ is the polar of $Q$ and $HG$ is the polar of $D$, hence $QQ\cap GH=\{T\}$ lies on the polar of $X$ and similarly, $PP\cap EF=\{S\}$ lies on the polar of $X$. Therefore it suffices to prove that $M$, $S$ and $T$ are collinear. Note that since $AC$ is the polar of $M$, we have that $M$, $P$ and $Q$ are collinear by Brokard. Pascal on $QFGPHE$ gives $I$, $J$ and $M$ collinear and Pascal on $QQFGHP$ gives $T$, $I$ and $M$ collinear, hence $T$ lies on $\overline{M-I-J}$. We can similarly deduce that $S$ lies on $\overline{M-I-J}$, therefore $\overline{M-S-T}$. $\blacksquare$