woo..w! very-very easy problem. There exists $E$ on $BC$ if and only if $ \angle A+\angle D \ge 180^\circ$. Analoguously, also point $F$. Hence, $E$ is exists, there exists $F$ on the side $AD$, if and only if $ \angle A+\angle D= \angle B+\angle C=180^\circ $ $ \Rightarrow $ $AB \parallel CD$.

sorry, $ \angle A+\angle D\le 180^\circ. $

mathuz wrote: woo..w! very-very easy problem. There exists $E$ on $BC$ if and only if $ \angle A+\angle D \ge 180^\circ$. Analoguously, also point $F$. Hence, $E$ is exists, there exists $F$ on the side $AD$, if and only if $ \angle A+\angle D= \angle B+\angle C=180^\circ $ $ \Rightarrow $ $AB \parallel CD$. actually $E$ is already determined by $A,B,D$ and $\angle ABC$ but $C$ is also so it's not iff. Anyway to solve this problem use the following lemma for part if(part only if is done by using the if part and point arrangements) For convex trapezoid $ABCD$ $AB||CD$ and $BC\cap AD=P$ ( $P-D-A$) the tangent from $A$ to $P$-excircle of $PDC$ and the tangent from $D$ to the incircle of $PAB$ are parallel. To prove this consider the centers of these 2 circles and some similar triangles.

We use the following lemma: Given two circles $\omega_1$ and $\omega_2$ and two points $E,F$ lying on the two different common external tangents of the circles, let the second tangents from $E,F$ to $\omega_1$ intersect at $P$, and similarly $Q$ for $\omega_2$. Then $PQ$ passes through the insimilicenter $Si$ of the two given circles. Indeed, by Monge's Theorem, it suffices to check that there exists a circle tangent to the four lines $FP, FQ, EP, EQ$, which is just a matter of segment length chasing. Back to the main problem: First suppose that such a point $F$ exists. Let the tangency points of $AB, CD$ with the respective inscribed circles be $R, S.$ Then by Brianchon's theorem, $R$ and $S$ both lie on $PQ$, where $P,Q$ are defined similarly as in the lemma. But $Si \in PQ$, i.e. $R,S$ are mapped to one another by the internal homothety, hence $AB \parallel CD.$ Conversely, if $AB \parallel CD$, let $F$ be the intersection of the second tangent from $B$ to the circle inscribed to $AECD$ with line $AD$. Since $Si = AC \cap BD$, by Pappus' theorem $Q = AE \cap BF$, $P = DE \cap CF$ and $Si$ are collinear. Hence by the lemma $FC$ is indeed tangent to the circle inscribed to $ABED$, as desired.

Let $P=BC\cap AD$, circles $\omega_1$ and $\omega _2$, with centers $O_1$ and $O_2$ are the incircles of $ABED$ and $AECD$, respectively. Suppose the tangent from $C$ to $\omega_1$ intersects $AD$ at $F_1$ and the tangent from $B$ to $\omega_2$ intersects $AD$ at $F_2$. Firstly assume $AB||CD$. After some angle chasing, we get $PCO_2\sim PO_1A\implies \frac {PC}{PO_1}=\frac {PO_2}{PA}\implies PCO_1\sim PO_2A$ $\implies \angle CO_1P=\angle O_2AP$. But note that $\angle O_2AP=\frac 1 2 \angle EAP$ and $\angle O_1PC+\angle CO_1P$ $=\angle BCO_1=\frac 1 2 (\angle F_1PC +\angle CF_1P)\implies \angle CO_1P=\frac 1 2 \angle CF_1P$. Hence $\angle EAP=\angle CF_1P\implies AE||CF_1$. Similarly $DE||BF_2$. But note that $PF_1=\frac {PC\cdot PA}{PE}=\frac {PD\cdot PB}{PE}=PF_2\implies F_1=F_2$. Using the if part, only if part is very easy.

Proof: Let us denote by $I_1$ and $I_2$ the incenters of quadrilaters $ABED$ and $AECD$, respectively. We now want to show that $\angle I_1EB +\angle I_2EC < 180$. But this is equivalent to \[\dfrac{\angle DEB + \angle AEC}{2} < 180 \Leftrightarrow \angle DEA < 180.\] The latter one is obviously true, so now we can assume that quadrilaters $DI_2EC$ and $AI_1EB$ have disjoints interiors. Now, let $DI_2$ and $AI_1$ intersect at point $S$. Since we now that $\angle CI_2S + \angle BI_1S = 180$, we obtain that $\angle DSA \ge 90$. And from here we get that $\angle A + \angle D \le 180$. Now, if there exists a point $F$ on side $AD$ with the property from the initial statement, this would immediately imply the asked parallelism and thus, the direct implication of the problem is now solved.$\blacksquare$ For the inverse, we will prove the following property, a lemma conjectured in the posts below that states the following:"Given a trapezoid $ABCD$ with $AB \parallel CD$ and $AD$ and $BC$ intersecting in some point $P$, than the tagent line from point $D$ to the excircle of triangle $PAB$ is parallel to the one from $B$ to the incircle of triangle $PCD$." Proof of lemma: Denote the incircle's center and the excircle's-as they are defined in the lemma by circles $O_1$ and $O_2$ respectively. Let the tangent point between $1$(the circle with center in $O_1$) and line $CD$ be $X$ and the the tangent from $B$ to $1$, intersect the latter at point $Y$. Denote by $Z$ the tangent point between $2$(the circle with center in $O_2$) and let the parallel line in $Z$ to $XY$ intersect $2$ again at point $R$. The common external tangents of $1$ and $2$ intersect them at $M, N, P, Q$ such that $M$ and $N$ lie on $1$ and $P$ and $Q$ on $2$. Now, by definition of point $R$, we know that it is the image of point $Y$ under the inverse homothety that sends $2$ to $1$. So we gain that $O_1X \parallel O_2Z, O_1Y \parallel O_2R, O_1M \parallel O_2Q$ and $O_1N \parallel O_2P$. Thus, by letting $M'$ and $N'$ to be the antipodes of $M$ and $N$ in $1$, respectively, we get that quadrilaters $XN'M'Y$ and $ZPQR$ are similar quadrilaters with sides respectively parallel. Thus, if we denote by $U$ and $V$ the intersections of the diagonals in the latter two, we get that $O_1U \parallel O_2V$. Now, by Pappus theorem for the hexagon $MN'XYM'N$ we get that $O_1, U$ and $W$ are collinear, where $W$ is the intersection of sides $MY$ and $NX$. But since $W$ is the pole of line $BD$ wrt. $1$, it means that $O_1U$ is perpendicular to $BD$ and therefor $O_2V$ is also perpendicular on line $BD$ and since $ZQ$ is the polar of point $B$ wrt. $2$, it follows that $DR$ is tangent to circle $2$. Since the inverse homothety that sends $1$ to $2$, also sends $Y$ to $R$, it means that the tangents in that points to the circle they are belonging to must be parallel and this ends the proof of the lemma.$\blacksquare$ And now, just observe that if one choose $F$ on the side $AD$ such that $CF \parallel AE $, then $BF \parallel DE$. But this is a very deep lemma which i will just conjecture as my folks did before...

wait how is this even possible? if ABED and ACED are cyclic then B=C or B=E etc. what is going on?

They are circumscribed, not cyclic. This may help you http://en.wikipedia.org/wiki/Tangential_quadrilateral

WLOG $AD$ and $BC$ meet at a point such that $X$ is closer to $B$ than $C$, and closer to $A$ than $D$. Let $\omega_1$ be the circle tangent to segments $BC, AD, AB$ and $\omega_2$ the circle tangent to $BC, AD, CD$. Note that $EA$ is tangent to $\omega_2$ and $ED$ is tangent to $\omega_1$. Take $F$ to be the intersection of the tangent from $B$ to $\omega_2$ and from $C$ to $\omega_1$. The claim is that $F\in BC \iff AB\parallel CD$. Let $Z_1=AE\cap BF$ and $Z_2=DE\cap CF$. Take $T_1$ to be the tangency point of $AB$ and $\omega_1$, and $T_2$ the tangency point of $CD$ and $\omega_2$. Suppose $EZ_1, FZ_1, AD, BC$ hit $\omega_2$ at $V_1, U_1, V_2, U_2$ and the incircle of $\triangle ABZ_1$ hits $AZ_1$ at $U_3$ and $BZ_1$ at $V_3$. We claim that independent of $F$ being on $AD$, the incircle of $ABZ_1$ is tangent to $AB$ at $T_1$. Note that $XU_2=XV_2$. Thus, $XB+BU_2=XA+AV_2$. But $BU_2=BV_1=BZ_1+Z_1V_1$ and $AV_2=AZ_1+Z_1U_1$, so $XB+BZ_1=XA+AZ_1$ since $Z_1U_1=Z_1V_1$. Note that $T_1$ is the intersection of the $X-$ excircle of $XAB$ with $AB$, so $2BT_1=XA+AB-XB = AB + BZ_1 - AZ_1$, implying $T_1$ is the intersection of the incircle of $ABZ_1$ with $AB$. Note that $T_1$ is the exsimilicenter of the incircle of $ABZ_1$ and $\omega_1$, $Z_1$ is the insimilicenter of the incircle of $ABZ_1$ and $\omega_2$, and take $S$ to be the insimilicenter of $\omega_1, \omega_2$. By Monge, it follows $T_1, Z_1, S$ are collinear. By a similar argument, $T_2, Z_2, S$ are collinear. Note this holds regardless of $F$ lying on $AD$ or not. If $F\in AD$ then note by Brianchon on $T_1BEZ_2FA$ we have $Z_1\in T_1Z_2$, so it follows that $T_1, Z_1, T_2, Z_2, $ and $S$ are collinear. But then a homothety at $S$ maps $T_1$ to $T_2$ and thus maps $AB$ to $CD$, implying $AB\parallel CD$ as desired. If $AB\parallel CD$, then a homothety maps the two. Because they are tangent to $\omega_1, \omega_2$, it follows the homothety is actually centered at $S$; i.e. $T_1$ maps to $T_2$ and $T_1, T_2, S$ are collinear. But then it follows $T_1, Z_1, T_2, Z_2, S$ are collinear again. But then by the converse of Brianchon on $T_1BEZ_2FA$, we have $AF$ is tangent to $\omega_1$, and thus $F\in AD$ as desired. $\square$

Here is an amusing solution (in particular to the "only if" part). First, we define a few points. Let $AD\cap BC$ and suppose that $A$ lie between $P$ and $D$. Let $\omega_1, \omega_2$ be the $P$-excircle of $\triangle PAB$ and the incircle of $\triangle PCD$ respectively. If point $E$ exists, note that incircles of $\square ABED$ and $\square AECD$ must be $\omega_1$ and $\omega_2$ respectively. Let $\omega_1$ touches $AB$ at $X$ and $\omega_2$ touches $CD$ at $Y$. If part Suppose that $AB\parallel CD$. Then since intouch point and ex-touch point are isotomic, we get $AX : XB = CY : YD$ or $AC, BD, XY$ are concurrent at $P$. Let $K=AE\cap XY$ and $L=DE\cap XY$. By the converse of Pappus' theorem, $BK,CL,AD$ are concurrent at say $F$. Then the converse of Brianchon's theorem on $FKECYD$ gives $FK$ is tangent to $\omega_2$. Similarly $FC$ is tangent to $\omega_1$ so we are done. Only if part Let $K=AE\cap BF$ and $L=DE\cap CF$. Then by Brianchon's theorem on $FKECYD$ and $FLEBXA$, we get $K,L,X,Y$ are colinear. By Pappus' theorem, we get $AC, BD, XY$ are concurrent at $Q$. We claim that this condition alone suffices to imply that $AB\parallel CD$. Ignore the existence of $E,F$ for a while. Let us fix points $P,C,D,Y$ and animate $AB$ so that it's parallel to a fixed direction. Clearly $X$ move linearly through $P$ hence it has degree $1$, which means $XY$ also has degree $1$. Now lines $AC, BD$ both have degree $1$. Therefore the concurrency of $AC,BD,XY$ is a polynomial equation of degree $3$. However, When $P=A=B$, we trivially get the concurrency. When $AB$ is tangent to $\omega_2$, which has two such points. We get by Newton's theorem that $AC,BD,XY$ are concurrent. Hence we get two more solutions. Therefore we get three solutions. But suppose that there is a fourth case, then this polynomial equation is trivial and holds for all positions of $A$. Now take $A=\infty_{PC}$ and $B=\infty_{PD}$. Then $AC\cap BD$ is the point $Q$ which $\square PCQD$ is parallelogram and we get that $P$-Nagel cevian of $\triangle PAB$ is parallel to $QY$. However, if $Y'$ be the ex-touch point of $\triangle PCD$ to $CD$. Then clearly $PY'\parallel QY$. Therefore $P$-Nagel cevian of $\triangle PAB$ and $\triangle PCD$ are parallel. Hence this forces $AB\parallel CD$ as desired.

StanleyST wrote: Proof: Let us denote by $I_1$ and $I_2$ the incenters of quadrilaters $ABED$ and $AECD$, respectively. We now want to show that $\angle I_1EB +\angle I_2EC < 180$. But this is equivalent to \[\dfrac{\angle DEB + \angle AEC}{2} < 180 \Leftrightarrow \angle DEA < 180.\]The latter one is obviously true, so now we can assume that quadrilaters $DI_2EC$ and $AI_1EB$ have disjoints interiors. Now, let $DI_2$ and $AI_1$ intersect at point $S$. Since we now that $\angle CI_2S + \angle BI_1S = 180$, we obtain that $\angle DSA \ge 90$. And from here we get that $\angle A + \angle D \le 180$. Now, if there exists a point $F$ on side $AD$ with the property from the initial statement, this would immediately imply the asked parallelism and thus, the direct implication of the problem is now solved.$\blacksquare$ In the first part of the proof, how does one obtain $\angle DSA\geq 90^{\circ}$?

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Suppose $F,E$ are points on the sides $JA$ and $JB$ of a triangle $\triangle JAB$. Let $H=AE\cap FB$, let $\omega$ be the incircle of $AJB$ and suppose it touches $AB$ at $T$. Suppose the second tangent of $F$ to $\omega$ intersect $BJ$ at $C$, and the second tangent of $E$ to $\omega$ intersect $AJ$ at $D$. Let $FC\cap DE=G$, $AC\cap DB=I$. If $HFJE$ is a circumscribed quadrilateral, then $T,H,I,G$ are collinear. Proof. By Pappus Theorem, $H,I,G$ are collinear. We now show that $T,H,I$ are collinear. Suppose $\omega$ touches $JA,JB$ at $M,N$. From Claim 3 of my post here, the incircle of $\triangle AHB$ is tangent to $AB$ at $T$. Now Suppose $AB\cap MN=Q$, $TH\cap MN=R$, we have $$-1=(Q,T;A,B)\overset{H}{=}(Q,R;P,O)$$Therefore, $R$ lies on $IH$, which proves the lemma. $\blacksquare$ We now return to the original problem, suppose $AD\cap BC=J$ WLOG assume $AB$ lies on different sides of $CD$ as $J$. Suppose such point $F$ exists, Suppose the incircle of $\triangle GCD$ intersect $CD$ at $U$, then the $J-$excircle of $\triangle JCD$ also intersect the line $CD$ at $U$. From the lemma, $T,H,I,G,U$ are collinear. Therefore, the homothety at $G$ will send $\omega$, the incircle of $\triangle JAB$ to $\omega_1$, the incircle of $\triangle GCD$, hence tangent at $T$ to $\omega$, AB and tangent at $U$ to $\omega_1$, CD are parallel as desired. Suppose $AB\|CD$, suppose the second tangent at $B$ to the $J-$excircle of $\triangle JCD$ intersect $AJ$ again at $F'$. Suppose the second tangent at $F$ and $C$ to incircle of $\triangle JAB$ intersect DE at $G_1,G_2$ respectively. Then the homothety at $G_2$ which sends the incircle of $\triangle JAB$ to the incircle of $\triangle G_2CD$ will sends $T$ to $U$, hence $T,G_2,U$ are collinear. The homothety at $H'$ with sends the inircle of $\triangle HAB$ to the incircle of $HF'JE$ also sends $T$ to $U$, this implies $T,H',U$ are collinear. Meanwhile, by the lemma, $T,H',G_1$ are collinear. Therefore, $G_1=TU\cap DE=G_2$ Therefore, $F'C$ is tangent to the incircle of $\triangle JAB$. Therefore, $AF'CB$ is circumscribed as well, hence $F'$ is the desired point, this completes the proof.

Denote by $\omega_1,\omega_2$ circles with centers $I_1,I_2$ which tangent to sides $DA,AB,BC$ and $BC,CD,DA$ of quadrilateral respectively. Redefine $E\in BC$ as intersection of tangents from $A$ to $\omega_2$ and from $D$ to $\omega_1$ different from $AB,CD.$ Redefine $F$ analogously. Assume that $AB\parallel CD.$ We claim that $AE\parallel CF$ and similarly $BF\parallel DE.$ Let $G=AD\cap BC,$ so $I_1I_2$ bisects angle $AGB$ and $$\measuredangle AI_1G=\frac{\measuredangle ABG}{2}=\frac{\measuredangle DCG}{2}=\measuredangle DCI_2\implies GAI_1\stackrel{+}{\sim} GI_2C\implies GAI_2\stackrel{+}{\sim} GI_1C\implies$$$$\implies \measuredangle BAE+\measuredangle FCD=\measuredangle BAD+\measuredangle BCD+2(\measuredangle DAI_2+\measuredangle I_1CB)=\measuredangle BGA+\measuredangle AGB=0\implies AE\parallel CF.$$Now Pappus on $(B,E,C),(\infty_{AE},\infty_{AB},\infty_{BF})$ gives $F\in AD$ $\Box$ Assume that $F\in AD.$ Let $X=AE\cap BF,$ $Y=CF\cap DE,P=\omega_1\cap AB,Q=\omega_2\cap CD.$ Easy to see that $$|EX|+|EY|=|FX|+|FY|,$$so there exist circle which tangent to all sidelines of $EXYF$ and by Monge center of internal homothety $\phi :\omega_1\mapsto \omega_2$ lies on $XY.$ Brianchon on $APBEYF,CQDFXE$ gives $P,Q\in XY,$ so in fact $\phi (P)=Q\implies AB\parallel CD$ $\Box$

Such a degen problem... too much projective. Let $X = BC \cap AD$ and let $\Gamma_1, \Gamma_2$ be the $X$-excircle of $XBC$ and incircle of $XAD$ respectively with WLOG $X$ is closer to $BC$ than $AD$. Then given a point $S$ on $BC$, we can draw a tangent from $S$ to either circle $\Gamma$ that is not $BC$ and intersect it with $AD$ to obtain a point on $AD$. We can follow this up with a projection at the point at infinity perpendicular to the angle bisector of $\angle BXC$ to yield a projective map from $BC$ to itself. The original map is projective because it is constituted by a projection from $AD$ to a pencil through the center of $\Gamma$, followed by a rotation via two composed reflections and ending with a projection from the pencil to $BC$. We now project all the points in the diagram via that point at infinity onto $BC$. Now let $g_1, g_2$ represent the two such maps via $\Gamma_1, \Gamma_2$. Align coordinate axes so that the angle bisector of $\angle BXC$ is vertical, then notice that top and bottom points of $\Gamma_1$, when projected horizontally onto $BC$, are fixed by $g_1$. Thus $g_1$ is an inversion through the horizontal projection of the center of $\Gamma_1$ onto $BC$. The same logic holds for $g_2$. Now that since we must have $\Gamma_1, \Gamma_2$ intersect, it follows that the two circles of inversion of $g_1, g_2$ also intersect. Let them intersect at $P, Q$, then there exists a circle centered at the exsimilicenter of the two circles of inversion passing through $P, Q$ that swaps the two circles of inversion. By scaling symmetry at $X$, it follows that the exsimilicenter of the two circles of inversion is $X$. Now take a homography mapping $P$ to $0$ and $Q$ to $\infty$, and scale so that $BC$ maps to the unit circle. Then the circles of inversion for $g_1, g_2$, and the circle around $X$ are each mapped to lines through the origin. Now notice that the existence of $F$ is equivalent to the claim that $g_1g_2g_1g_2g_1g_2 = e$, or that the order of $g_1g_2$ is exactly three. In this case, since $g_1g_2$ is a rotation around the origin, it must be $120^{\circ}$, so it follows that two lines corresponding to $g_1, g_2$ are $60^{\circ}$ apart. Finally, since these are symmetric around $x$, the line corresponding to the circle through $X$, it follows that that line is either their internal angle bisector or their external angle bisector. If the angles between these lines are $30^{\circ}$, since $E, F$ must lie on their respective sides, that means there must be a point on the unit circle in the length-six image of the $A$ both between $x, g_1$ and $x, g_2$. However this is impossible since the sextant which the point lies in rotates each time $g_1, g_2$ are taken in their described order, so it must be the case that $g_1, g_2, x$ are pairwise $60^{\circ}$ from one another. Now one can easily verify that a reflection about $g_1, g_2, g_1$ is equivalent to a reflection about $x$, so it follows that $C, D$ are the images of $A, B$ under an inversion at $X$ after the projection onto $BC$, so in the original diagram, it follows that $AB$ is parallel to $CD$. If you start from $AB || CD$, you get from algebra that since $g_1, g_2$ are symmetric about $x$, and $g_1g_2g_1 = x$, you get that $g_1$ is one of the angle bisectors between $x, g_2$ and already know that $x$ is one of the angle bisectors between $g_1, g_2$. It follows that they are at $60^{\circ}$ angles from one another, and then you can derive the original order $3$ from there. Thus we have proven both directions, finishing.

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First, we prove a lemma. Lemma wrote: Let $ABCD$ be a non-tangential quadrilateral such that $AD$ is not parallel to $CB$. Let $AD$ intersect $BC$ at $E$, $AC$ intersect $BD$ at $M$. Let $Y$ be the point where the $E$-excenter of $\triangle ECD$ intersects $CD$, and let $Z$ be the point where the incenter of $\triangle EAB$ intersects $AB$. Then, $Z$, $M$, $Y$ are collinear if and only if $AB\parallel CD$. If $AB\parallel CD$ then let $Z'$ be the point on $AB$ such that $AZ=BZ'$. Note that $Z'$ is the point where the $E$-excenter of $\triangle EAB$ intersects $AB$, so the homothety at $E$ that takes $CD$ to $AB$ takes $Y$ to $Z'$, so $E$, $Y$, $Z'$ are collinear. Let $YM$ intersect $AB$ at $Y'$ then \[(P_\infty,Z;B,A)=(P_\infty,Z';A,B)\stackrel{E}{=}(P_\infty,Y;D,C)\stackrel{M}{=}(P_\infty,Y';B,A)\]so $Z=Y'$ and so $Z$, $M$, $Y$ collinear. Now suppose $Z$, $M$, $Y$ are collinear. If $AB$ and $CD$ are parallel then we are done. Otherwise, let them intersect at $P$. Without loss of generality, let $A$ and $D$ be closer to $P$, $C$ and $D$ be closer to $E$. We have $(P,Z;A,B)\stackrel{M}{=}(P,Y;C,D)$ which gives \[\frac{PA\cdot ZB}{ZA\cdot PB}=\frac{PC\cdot YD}{YC\cdot PD}\implies \frac{PA\cdot PD}{PB\cdot PC}=\frac{ZA\cdot YD}{YC\cdot ZB}\]Note that \begin{align*} \frac{PA\cdot PD}{PB\cdot PC} &= \frac{PA}{PC}\cdot \frac{PD}{PB}\\ &= \frac{\sin(\angle PCA)}{\sin(\angle PAC)}\cdot \frac{\sin(\angle PBD)}{\sin(\angle PDB)}\\ &= \frac{\sin(\angle DCM)}{\sin(\angle BAM)}\cdot \frac{\sin(\angle ABM)}{\sin(\angle CDM)}\\ &= \frac{\sin(\angle DCM)}{\sin(\angle CDM)}\cdot \frac{\sin(\angle ABM)}{\sin(\angle BAM)}\\ &= \frac{DM}{CM}\cdot \frac{AM}{BM}\\ &= \frac{MA}{MC}\cdot \frac{MD}{MB} \end{align*}We now use Menelaus to further manipulate this. We know that from Menelaus on $\triangle ACE$ and line $MBD$: \[\frac{MA}{MC}\cdot \frac{BC}{BE}\cdot \frac{DE}{DA}=1\]and from Menelaus on $\triangle BDE$ and line $MAC$: \[\frac{MD}{MB}\cdot \frac{AE}{AD}\cdot \frac{CB}{CE}=1\]so \[\frac{MA}{MC}\cdot \frac{MD}{MB}=\frac{AD^2\cdot BE\cdot CE}{BC^2\cdot AE\cdot DE}\]Let $a,b,c,d$ denote $EA$, $EB$, $EC$, and $E$, respectively. Let $x$ be $AB$ and $y$ be $CD$. From the Law of Cosines, we have \[\frac{x^2-a^2-b^2}{2ab}=\frac{y^2-c^2-d^2}{2cd}=-\cos(\angle E)\]so \[\frac{x^2-a^2-b^2+2ab}{2ab}=\frac{y^2-c^2-d^2+2cd}{2cd}\]Note that $x^2-a^2-b^2+2ab=(x-a+b)(x+a-b)=4AZ\cdot BZ$ and $y^2-c^2-d^2+2cd=(y-c+d)(y+d-c)=4CY\cdot DY$ so \[\frac{AZ\cdot BZ}{CY\cdot DY}=\frac{x^2-a^2-b^2+2ab}{y^2-c^2-d^2+2cd}=\frac{ab}{cd}\]Note that the second fraction cannot be indeterminate as that would imply $y=c-d$ which violates the triangle inequality. Now, \[\frac{AZ\cdot DY}{CY\cdot BZ}=\frac{AD^2bc}{BC^2ad}\]and dividing the two and taking the square root gives that \[\frac{DY}{BZ}=\frac{AD\cdot CE}{BC\cdot AE}\implies \frac{DA}{DY}\cdot \frac{EC}{EA}\cdot \frac{BZ}{BC}=1\]recall that \[\frac{MB}{MD}\cdot \frac{AD}{AE}\cdot \frac{CE}{CB}=1\]from Menelaus and so we can compare the two and deduce that \[\frac{MB}{MD}=\frac{BZ}{DY}\]and this implies that $\angle BZM$ and $\angle MYD$ are either equal or supplements. Since we already assumed that $AB$ isn't parallel to $CD$, we can assume that they are supplements. Let the altitudes from $Z$ to $AB$ and from $Y$ to $CD$ intersect at $O$. Since $\angle BZY=\angle CYZ$, we have $\triangle YOZ$ isosceles so $OZ=OY$. Therefore, there exists a circle $\omega$ centered at $O$ tangent to $AB$ at $Z$ and $CD$ at $Y$. Let $A'$, $B'$, $C'$, $D'$ be on $AZ,BZ,CY,DY$, respectively, such that $A'D'$ is parallel to $AD$, $B'C'$ is parallel to $BC$, and $\omega$ is the incircle of $A'B'C'D'$. There is a homothety centered at $Z$ that takes the incircle of $\triangle EAB$ to $\omega$. This homothety must also take $AB$ to $A'B'$. In particular, we have $A'Z/AZ=B'Z/BZ$. Similarly, $C'Y/CY=D'Y/DY$. Let $PZ=PY=z$ because $\angle PZY=\angle PYZ$, and let $AZ=p$, $BZ=q$, $DY=r$, and $CY=s$. Furthermore, let $A'Z=pu$, $B'Z=qu$, $D'Y=rv$, and $C'Y=sv$. The factors of the two homotheties taking the incircle of $\triangle EAB$ and the $E$-excircle of $\triangle ECD$ to $\omega$. are $u$ and $v$, respectively. We have $AD\parallel A'D'$ and $BC\parallel B'C'$ implying \[\frac{z-p}{z-pu}=\frac{z-r}{z-rv}\text{ and }\frac{z+q}{z+qu}=\frac{z+s}{z+sv}\]By Brianchon's on $A'ZB'C'YD'$, $A'C'$, $B'D'$ and $YZ$ are concurrent. Now note that $(P,Z;A,B)\stackrel{M}{=}(P,Y,C,D)$ and $(P,Z;A',B')\stackrel{M}{=}(P,Y,C',D')$ gives \[\frac{(z-p)q}{(z+q)p}=\frac{(z+s)r}{(z-r)s}\text{ and }\frac{(z-pu)q}{(z+qu)p}=\frac{(z+sv)r}{(z-rv)s}\]Combining the two gives \[\frac{z-p}{z-pu}\cdot \frac{z+qu}{z+q}=\frac{z+s}{z+sv}\cdot \frac{z-rv}{z-r}=\frac{z-pu}{z-u}\cdot \frac{z+q}{z+qu}\]and this implies that $PA/PB=PA'/PB'$, which implies that $u=1$. This also implies $v=1$ so $\omega$ is both the incircle of $\triangle EAD$ and the $E$-excircle of $\triangle CDE$, implying that $ABCD$ is tangential, which we assumed is not true. This concludes our proof of the lemma. $\blacksquare$ $~$ Now we proceed with the main argument. Let $AD$ and $BC$ intersect at $W$. Without loss of generality, let $C$ and $D$ be closer to $W$. Let the incircle of $\triangle ABW$ be $\omega_1$ and intersect $AB$ at $Z$. Let the $W$-excircle of $\triangle CDW$ be $\omega_2$ and intersect $CD$ at $Y$. Let $AC$ and $BD$ intersect at $M$. Since $ABED$ and $AECD$ are tangential, $AE$ is tangent to $\omega_2$ and $ED$ is tangent to $\omega_1$. Let $F$ be the point such that $CF$ is tangent to $\omega_1$. A point as described in the problem exists if and only if $BF$ is tangent to $\omega_2$. Let $AE$ and $BF$ intersect at $T$, $CF$ intersect $DE$ at $U$. By Brianchon's on $AZBEUF$, $Z$ is on $TU$. By Pappus on $AFD$ and $BEC$, $M$ is on $TU$. $Y$ is on $TU$ if and only if $Y$ is on $ZM$. Since $ABCD$ is clearly not tangential, by the lemma, this is if and only if $AB$ and $CD$ are parallel. Therefore, $AB\parallel CD$ if and only if $Y$ is on $TU$. Clearly, if $BF$ is tangent to $\omega_2$ then Brianchon's on $FTECYD$ gives $Y$ on $TU$, which in turn gives $AB\parallel CD$. Suppose $AB\parallel CD$. Redefine $F$ to be the intersection of the tangents from $B$ to $\omega_2$ and $C$ to $\omega_1$. We'll show that $F$ is on $AD$ and thus is the point we want. There exists a negative homothety at $M$ that takes $AB$ to $CD$. This also takes $\omega_1$ to $\omega_2$ so it will take line $BF$ to line $DE$. In particular, $BF\parallel DE$ and so $AE\parallel CF$. Let $AE$ and $DE$ intersect $CD$ and $AB$ at $H$ and $I$, respectively. Let $BF$ and $CF$ intersect $CD$ and $AB$ at $J$ and $K$ respectively. Note that \[(P_\infty,B;A,I)\stackrel{E}{=}(P_\infty,C;H,D)\implies BI\cdot CH=AB\cdot CD\]but we see that $BIDJ$ and $KAHC$ are parallelograms so $DJ=BI$ and $KA=CH$. This means that $JD\cdot KA=AB\cdot CD$ so $(P_\infty,A;B,K)=(P_\infty,D;J,C)$. This means that $AD$, $BJ$, and $CK$ are concurrent which means that $F$ is on $AD$ as desired.