Solution: Let $M, J$ be the midpoints of $CD, AB$, respectively; $K, T$ be the projections of $P$ on $AB, CD$. The line through $E_1$ and perpendicular to $CD$ intersects $PK$ at $I$, the line through $E_2$ and perpendicular to $AB$ intersects $PT$ at $L$. $JE_1\cap PK=\{Q\}, ME_2\cap PT=\{R\}; AB\cap CD=\{G\}$. Since $PK//E_2H$ we get $\angle E_2LH_2= \angle TPK$. Similarly, $\angle E_1IH_1=\angle TPK$. So $\angle E_2LH_2=\angle E_1IH_1$. On the other side, we will show that $\angle RE_2M=\angle JE_1Q$ $\Leftrightarrow \angle TLE_2+\angle LRE_2=270^o-\angle E_1JK-\angle E_1IH_1$ $\Leftrightarrow 2\angle TPK=90^o-\angle E_1JK+\angle TRM=\angle JQK+\angle TRM$ $\Leftrightarrow 2\angle G=\angle O_1PH_1+\angle O_2PH_2=|\angle A-\angle B|+|\angle C-\angle D|$ (right) Therefore two triangles $E_2LR$ and $E_1QI$ have $\angle RE_2L=\angle QE_1I$ and $\angle RLE_2+\angle QIE_1=180^o.$ Applying the law of sine we claim $\frac{QI}{E_1Q}=\frac{RL}{E_2L}$ or $\frac{QI}{\frac{1}{2}R_1}=\frac{RL}{\frac{1}{2}R_2}$ But $\frac{O_1J}{O_2M}=\frac{R_1\cos P}{R_2\cos P}=\frac{R_1}{R_2}$ then $\frac{QI}{RL}=\frac{O_1J}{O_2M}=\frac{PH_1}{PH_2}$. So $\frac{IH_1}{PI}=\frac{PL}{LH_2}$. Applying Thales's theorem we are done.

Easy to see that DCO_2 ~ ABO_1 , so O_2P/O_1P = H_2P/H_1P and angle O_2PO_1 = H_1PH_2 , so O_1O_2P ~ H_2H_1P . Let P' is on E_1P and P'P_1 = E_1P , and let G is on E_1H_2 and GE_1 = E_1H_2 , let P'G intersect O_1P at point I and P'G intersect H_1P at point F and H_1H_2 at point K . Easy to see that FKH_1 ~ P'GO_1 ~ O_1O_2P ~ H_2H_1P , and KP'H_1 ~ O_1JO_2 , so JO_2/O_2P = P'K/KF and P'J || FP ,so KO_2 || FP , let M is midpoint OF KH_2 , easy to see that ME_2 || KO_2 || H_1P and ME_1 || KG || H_2P , so done

Let $p=0$, $r^2=a/\overline{a}=d/\overline{d}$, and $s^2=b/\overline{b}=c/\overline{c}$ such that $r^2\ne s^2$. Letting $x$ denote the unique point such that $xe_1\perp dc$ and $xe_2\perp ab$, we easily get $h_1=\frac{(r^2+s^2)(a-b)}{r^2-s^2}$, $h_2=\frac{(r^2+s^2)(d-c)}{r^2-s^2}$, $e_1=\frac{o_1+h_1}{2}=\frac{a+b+h_1}{4}=\frac{ar^2-bs^2}{2(r^2-s^2)}$, $e_2=\frac{dr^2-cs^2}{2(r^2-s^2)}$, and \begin{align*} xe_2\perp ab \implies -2r^2s^2(r^2-s^2)\overline{x}+cr^4-ds^4 &= \frac{(as^2-br^2)[2(r^2-s^2)x+cs^2-dr^2]}{a-b} \\ xe_1\perp dc \implies -2r^2s^2(r^2-s^2)\overline{x}+br^4-as^4 &= \frac{(ds^2-cr^2)[2(r^2-s^2)x+bs^2-ar^2]}{d-c} \\ \end{align*}whence subtracting the two equations and factoring to cancel out $r^2-s^2$ yields \begin{align*} x &= \frac{(r^2+s^2)[ac(a-c)+bd(b-d)]+(ad+bc)[(c-b)r^2+(d-a)s^2]}{2(r^2-s^2)(ac-bd)} \\ &= \frac{r^2[ac(a-c)+bd(b-d)+(ad+bc)(c-b)]+s^2[ac(a-c)+bd(b-d)+(ad+bc)(d-a)]}{2(r^2-s^2)(ac-bd)} \end{align*}By standard factoring techniques (e.g. considering an expression as a quadratic in one variable) \begin{align*} x-h_1 &= \frac{(a-b+c-d)[b(c+d)-ac]r^2-(a-b+c-d)[a(c+d)-bd]s^2}{2(r^2-s^2)(ac-bd)} \\ x-h_2 &= \frac{(a-b+c-d)[c(a+b)-bd]r^2-(a-b+c-d)[d(a+b)-ac]s^2}{2(r^2-s^2)(ac-bd)}, \end{align*}so \[\frac{x-h_1}{x-h_2}=\frac{bcr^2-ads^2+(bd-ac)(r^2+s^2)}{bcr^2-ads^2+(ac-bd)(r^2+s^2)}\in\mathbb{R},\]as desired.

I used a very natural and straightforward approach. I think there may be shorter solutions, but this one is very natural.

With this approach the problem is very simple, but it requires a lot of minor computations and work.

Let $Q$ be such that $XE_1QE_2$ is a parallelogram. Now, let $O_1$, $O_2$, $H_1$, $H_2$ be vectors. We have that $X=E_1+E_2-Q$. Our result is equivalent to proving that $H_1$, $H_2$, and $\frac {O_1+O_2+H_1+H_2-P-J} {2}$ are collinear, since $2Q=P+J$. Note that $JQ$ is where the perpendicular from $E_2$ to $CD$ meets that from $E_1$ to $AB$, so $J$ is where that from $O_2$ to $CD$ meets that from $O_1$ to $AB$. Thus $JA=JB$ and $JC=JD$. Now fix $C$, $D$, and the orientation of $AB$. Then vary $AB$ and have $A$ move linearly. We have that $H_2$ stays put, but $H_1$ moves in a direction perpendicular to $AB$ (whose direction is fixed). The last term has everything fixed except $\frac {H_1+O_1-J} {2}$. We have $H_1$ moves in a direction perpendicular to $AB$ and that $O_1-J$ is always perpendicular to $AB$ since $O_1$, $J$ are on its perpendicular bisector, so that can only move in a direction perpendicular to $AB$. Also everything moves linearly by homothety. So it suffices to prove this for two positions of point $A$, because the point where the line between $H_1$, $H_2$ meets the line on which $\frac {O_1+O_2+H_1+H_2-P-J} {2}$ varies is linear with $A$, as is $\frac {O_1+O_2+H_1+H_2-P-J} {2}$ itself. For one, choose $A=D$, and for the other choose $A=A'$ to force $B=C$. These are analogous, so we will just do the $A=D$ case, and this suffices. Now, where $A=D$ note $JA=JB=JC$. Now $O_1+O_2 +H_1-H_2$ $O_1+O_2 -H_1+H_2$, and $P+J$ are just $H_2$, $H_1$, $X$ reflected over the midpoint of $E_1E_2$ and then scaled up by a homothety of $2$. Let $PCB$ be the x-axis. Now, the line defined by $O_1+O_2+k(H_1-H_2)$ is just the line of points whose $x$ component is $\frac {2p+b+c} {2}$, because $H_1H_2$ is a line perpendicular to $BC$ and $\frac {2p+b+c} {2}$ is the sum of coefficients of $x$ coordinates. Now, $P+J$ has $x$ coordinate $\frac {2p+b+c} {2}$ because $J$ is just the circumcenter of $ABC=DBC$. That's it!

Let's concentrate to $\triangle AO_2H_2$, now $AO_1$ and $AH_2$ are isogonal conjugates (obvious )! Let $H$ be the reflexion of $H_1$ through $AH_2$, now $\frac{AO_1}{AH_1}=\frac{AO_2}{AH_2}= \frac{1}{2} sec \angle A \implies \triangle AO_1O_2 \sim \triangle AH_1H_2 \sim \triangle AHH_2$ and also the spiral similarity with center $A$ sends $\triangle AO_1O_2$ to $\triangle AHH_2$. now the perpendicular from $E_1$ to $CD$ and $E_2$ to $AB$ are, respectively parallel to $AH_2$ and $AH_1$. so if the 1st perp. bisector intersects $H_1H_2$ at $P$, it suffices to prove that $PE_2 \parallel AH_1$. Define $X = AH_2 \cap O_1H, Y = AH_2 \cap O_1H_1, Z = AH_1 \cap O_2H_2, M = HH_1 \cap AH_2$, i.e., $M$ is the midpoint of $HH_1$. Now $A$ is also the center of spiral similarity that takes $\triangle AO_2H_2$ to $\triangle AO_1H$ and also $\triangle AO_2Z$ to $\triangle AO_1X$, Thus $\frac{E_2Z}{ZH_2}= \frac{EX}{XH} [ E$ is the midpoint of $O_1H ]$. Now, $PE_2 \parallel AH_1 \iff \frac{PH_1}{H_1H_2} = \frac{E_2Z}{ZH_2} \iff \frac{E_1H_1}{H_1Y} = \frac{EX}{XH} \iff \frac{O_1X}{XH} = \frac{O_1Y}{H_1Y}$ $[ \because E_1$ and $E_2$ are, respectively, the midpoints of $O_1H_2$ and $O_2H_2 ]$ - which is obvious applying Menelaus' Theorem to $\triangle O_1H_1H$ and the line $\overline{XMY}$. Thus our proof is done! $[$ NB: My whole solution uses $A$ instead of $P$ but the proof remains the same $]$

This is a beautiful problem, mostly because it is quite troll. It becomes much simpler once one realises that points $A$, $B$, $C$ and $D$ are absolutely irrelevant. Specifically, let $\theta=\angle{APD}$. Then it's well known that $$\frac{PH_1}{PO_1}=\frac{PH_2}{PO_2}=2\cos\theta$$and moreover, since $O_1$ and $H_1$ are isogonal conjugates with respect to $ABP$ and $O_2$ and $H_2$ are isogonal conjugates with respect to $CDP$, $\angle{(PD, PH_2)}=\angle{(PO_2, PC)}$ and $\angle{(PA, PH_1)}=\angle{(PO_1, PB)}$, hence $\angle{H_2PH_1}=\angle{O_1PH_2}$. Hence $PO_1O_2\sim PH_1H_2$. And now it suffices to solve the following problem: Let $PO_1O_2$ and $PH_1H_2$ be similar triangles (with opposite orientations). Let $E_1$ and $E_2$ be the midpoints of $O_1H_1$ and $O_2H_2$, respectively. Prove that the line parallel to $PH_2$ through $E_1$, the line parallel to $PH_2$ through $E_2$ and line $H_1H_2$ concur. [asy][asy] import graph; size(14.65200000000002cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.860000000000006, xmax = 19.56000000000003, ymin = -3.560000000000005, ymax = 8.960000000000013; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921575,0.6274509803921575,0.6274509803921575); pen qqzzqq = rgb(0.000000000000000,0.6000000000000006,0.000000000000000); pen wwzzqq = rgb(0.4000000000000005,0.6000000000000006,0.000000000000000); draw((6.300000000000007,5.220000000000006)--(2.100000000000002,-2.060000000000002)--(-1.904342243116190,-2.064204701141710)--cycle, aqaqaq); draw((6.300000000000007,5.220000000000006)--(7.740000000000009,1.640000000000002)--(9.560000000000011,1.380000000000001)--cycle, qqzzqq); draw((11.77870355734198,5.225752832726087)--(7.578703557341980,-2.054247167273919)--(3.574361314225789,-2.058451868415627)--cycle, aqaqaq); /* draw figures */ draw((6.300000000000007,5.220000000000006)--(2.100000000000002,-2.060000000000002), linewidth(1.200000000000002) + aqaqaq); draw((2.100000000000002,-2.060000000000002)--(-1.904342243116190,-2.064204701141710), linewidth(2.000000000000000) + aqaqaq); draw((-1.904342243116190,-2.064204701141710)--(6.300000000000007,5.220000000000006), linewidth(1.200000000000002) + aqaqaq); draw((6.300000000000007,5.220000000000006)--(7.740000000000009,1.640000000000002), linewidth(1.200000000000002) + qqzzqq); draw((6.300000000000007,5.220000000000006)--(9.560000000000011,1.380000000000001), linewidth(1.200000000000002) + qqzzqq); draw((9.560000000000011,1.380000000000001)--(7.740000000000009,1.640000000000002), linewidth(1.200000000000002) + qqzzqq); draw((11.77870355734198,5.225752832726087)--(3.574361314225789,-2.058451868415627), linewidth(1.200000000000002) + aqaqaq); draw((11.77870355734198,5.225752832726087)--(7.578703557341980,-2.054247167273919), linewidth(1.200000000000002) + aqaqaq); draw((2.100000000000002,-2.060000000000002)--(7.740000000000009,1.640000000000002), linetype("4 4")); draw((6.300000000000007,5.220000000000006)--(2.100000000000002,-2.060000000000002), aqaqaq); draw((2.100000000000002,-2.060000000000002)--(-1.904342243116190,-2.064204701141710), aqaqaq); draw((-1.904342243116190,-2.064204701141710)--(6.300000000000007,5.220000000000006), aqaqaq); draw((6.300000000000007,5.220000000000006)--(7.740000000000009,1.640000000000002), qqzzqq); draw((7.740000000000009,1.640000000000002)--(9.560000000000011,1.380000000000001), qqzzqq); draw((9.560000000000011,1.380000000000001)--(6.300000000000007,5.220000000000006), qqzzqq); draw((11.77870355734198,5.225752832726087)--(7.578703557341980,-2.054247167273919), aqaqaq); draw((7.578703557341980,-2.054247167273919)--(3.574361314225789,-2.058451868415627), aqaqaq); draw((3.574361314225789,-2.058451868415627)--(11.77870355734198,5.225752832726087), aqaqaq); draw(circle((9.040381033022875,4.242667231160039), 2.909444542726302), linewidth(1.600000000000002) + wwzzqq); draw((9.560000000000011,1.380000000000001)--(-1.904342243116190,-2.064204701141710), linetype("4 4")); draw((6.300000000000007,5.220000000000006)--(11.77870355734198,5.225752832726087)); draw((7.578703557341980,-2.054247167273919)--(-1.904342243116190,-2.064204701141710), linewidth(2.000000000000000) + aqaqaq); /* dots and labels */ dot((6.300000000000007,5.220000000000006),dotstyle); label("$P$", (6.380000000000014,5.340000000000008), NE * labelscalefactor); dot((7.740000000000009,1.640000000000002),dotstyle); label("$O_1$", (7.820000000000017,1.760000000000003), NE * labelscalefactor); dot((9.560000000000011,1.380000000000001),dotstyle); label("$O_2$", (9.640000000000020,1.500000000000002), NE * labelscalefactor); dot((2.100000000000002,-2.060000000000002),dotstyle); label("$H_1$", (2.180000000000006,-1.940000000000003), NE * labelscalefactor); dot((-1.904342243116190,-2.064204701141710),dotstyle); label("$H_2$", (-1.820000000000001,-1.940000000000003), NE * labelscalefactor); dot((4.920000000000005,-0.2100000000000003),dotstyle); label("$E_1$", (5.060000000000011,-0.3800000000000006), NE * labelscalefactor); dot((3.827828878441911,-0.3421023505708544),dotstyle); label("$E_2$", (3.900000000000010,-0.2200000000000003), NE * labelscalefactor); dot((11.77870355734198,5.225752832726087),dotstyle); label("$Q$", (11.86000000000002,5.340000000000008), NE * labelscalefactor); dot((7.578703557341980,-2.054247167273919),dotstyle); label("$X_1$", (7.660000000000017,-1.940000000000003), NE * labelscalefactor); dot((3.574361314225789,-2.058451868415627),dotstyle); label("$X_2$", (3.660000000000009,-1.940000000000003), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let the line through $O_1$ parallel to $PH_2$ and the line through $O_2$ parallel to $PH_1$ meet $H_1H_2$ at $X_2$ and $X_1$, respectively. Let $Q=O_1X_2\cap O_2X_1$. Since $QX_1$ is parallel to $PH_1$ and $QX_2$ is parallel to $PH_2$, triangles $PH_1H_2$ and $QX_1X_2$ are similar. Hence $\angle{O_1QO_2}=\angle{X_2QO_1}=\angle{H_2PH_1}=\angle{O_1PO_2}$; it follows that $PQO_2O_1$ is cyclic. Therefore, $\angle{PQO_2}=\pi-\angle{O_2O_1P}=\pi-\angle{PH_1H_2}$, hence lines $PQ$ and $H_1H_2$ are parallel. It easily follows that similar triangles $PH_1H_2$ and $QX_1X_2$ are actually congruent; hence $H_1H_2=X_1X_2$. Therefore, segments $H_2X_1$ and $H_1X_2$ have the same midpoint. Let $M$ be this common midpoint. Then $ME_1\parallel X_2O_1\parallel PH_2$ and $ME_2\parallel X_1O_2\parallel PH_1$, hence $M$ is the desired concurrency point, establishing the problem statement.

Here is a sketch of my solution:- (will add the details later on) By the Gauss Bodenmiller Line Theorem, we have that $H_1H_2$ is the radical axis of the circles with diameter $AC,BD,PQ$ respectively. Thus, $H_1H_2$ is the locus of all points $X$ such that $XA^2+XC^2-XB^2-XD^2=AC^2-BD^2$ and so, if we let $T$ be the intersection of the lines through $E_1,E_2$ perpendicular to $CD,AB$ respectively then we need only $T$ to satisfy this condition. Thus, we want $AE_2^2-BE_2^2+CE_1^2-DE_1^2=AC^2-BD^2$ and so, after using Appolonius formula, we are left with showing that \begin{align*} (AO_2^2-BO_2^2+CO_1^2-DO_1^2)+(AH_2^2-BH_2^2+CH_1^2-DH_1^2)=2(AC^2-BD^2) \end{align*} This is just bashing by the perpendicularity Lemma. (Will post the rest later) Thank v_Enhance's book which has the key idea of Gauss-Bodenmiller line.

Ignore this post.

Let $M,N,R$ be the midpoints of $AC,DB,BC$ respectively.By Newton-Gauss line we have $H_1H_2\perp MR$ so we want to show that $\triangle MNR$ and $\triangle H_2E_2E_1$ are orthologic,i.e by Carnot's it's sufficient that : $$H_2M^2-H_2N^2+NE_1^2-RE_1^2+E_2R^2-E_2M^2=0$$By noting that $H_2$ lies on radical axis of $(AC),(BD)$ we have that $H_2M^2-H_2N^2=\tfrac{BD^2}{4}-\tfrac{AC^2}{4}$.And the rest will be done by vectors set $\mathbf{P}$ to be the origin.$\mathbf{E_1}=\tfrac{\mathbf{A}+\mathbf{B}}{2}$ and $\mathbf{E_2}=\tfrac{\mathbf{C}+\mathbf{D}}{2}$ and so $NE_1^2=\left(\tfrac{\mathbf{A}-\mathbf{D}}{2}\right )^2$ and analogous for others.Everything now boils down on : $$(\mathbf{A}-\mathbf{D})^2-(\mathbf{A}-\mathbf{C})^2+(\mathbf{D}-\mathbf{B})^2-(\mathbf{D}-\mathbf{A})^2=(BD^2-AC^2)$$And hence we're done.$\blacksquare$

Let $M_1, M_2$ be midpoints of $ AB, CD$. Then notice $\triangle ABO_1\cup M_1\sim\triangle CDO_2\cup M_2$ so $$\frac{PH_1}{PH_2} = \frac{2O_1M_1}{2O_2M_2} = \frac{AO_1}{CO_2} = \frac{PO_1}{PO_2}$$Moreover, since $(PH_1, PO_1)$ and $(PH_2, PO_2)$ are isogonal wrt $\angle(AD, BC)$. Thus $\angle H_1PH_2=\angle O_1PO_2$ or $\triangle PH_1H_2\stackrel{-}{\sim}\triangle PO_1O_2$ Now let $Q$ be the reflection of $P$ across $H_1H_2$ and $M$ be the projection of $P$ onto $H_1H_2$. Since $\triangle QH_1H_2\stackrel{+}{\sim}\triangle PO_1O_2$. By spiral similarity, this triangle is also similar to $\triangle ME_1E_2$. Now let $H_1H_2$ meet $\odot(ME_1E_2)$ at $K$. Hence $$\angle H_1KE_1 = \angle ME_2E_1 = \angle PH_2H_1\implies KE_1\parallel PH_2\perp CD.$$Siimilarly $KE_2\perp AB$ so $K$ is the concurrency point and we are done.

See here

Here is another boring computational solution: Let $\mathcal P(P, \omega)$ be the power of $P$ wrt $\omega$. We use directed lengths. Let $O$ be the intersection of the lines through $O_1, O_2$ perpendicular to $CD$, $AB$, and let $H$ be the intersection of the lines through $H_1, H_2$ perpendicular to $CD$, $AB$. Then, the intersection of the lines through $E_1, E_2$ is the midpoint $E$ of $OH$. By Linearity of Power of a Point and since $H_1H_2$ is the radical axis of $(AC), (BD)$, it suffices to show that \[ \mathcal P(H, (AC)) + \mathcal P(O, (AC)) - \mathcal P(H, (BD)) - \mathcal P(O, (BD)) = 0. \]Recall that if $M$ is the midpoint of $XY$, then\[ \mathcal P(P, (XY)) = PM^2 - MX^2 = \frac{1}{2} (PX^2+PY^2-XY^2). \]So, we wish to show that \[ HA^2 + HC^2 - HB^2 - HD^2 + OA^2 + OC^2 - OB^2 - OD^2 = 2(AC^2-BD^2). \]Note that \[ HA^2 - HB^2 = H_2A^2-H_2B^2 = (AC^2-BD^2) - (PC^2-PD^2). \]where we used $H_2A^2 - AC^2 = PH_2^2 - PC^2$. Similarly $HC^2 - HD^2 = (AC^2-BD^2) - (PA^2-PB^2)$. Furthermore, \[ AO^2 - BO^2 = AO_2^2-BO_2^2 = \mathcal P(A, (PCD)) - \mathcal P(B, (PCD)) = AP\cdot AD - BP\cdot BC = PB\cdot BC - PA\cdot AD. \]Similarly, $CO^2 - DO^2 = PD\cdot DA - PC\cdot CB = BC\cdot PC - PD\cdot AD$. Note that we thus have\[ AO^2-BO^2 + CO^2 - DO^2 = BC(PB+PC) - AD(PA+PD) = (PC-PB)(PB+PC) - (PD-PA)(PA+PD) = PA^2+PC^2-PB^2-PD^2, \]thus we indeed have \[ HA^2 + HC^2 - HB^2 - HD^2 + OA^2 + OC^2 - OB^2 - OD^2 = (AC^2-BD^2) - (PC^2-PD^2) + (AC^2-BD^2) - (PA^2-PB^2) + PA^2+PC^2-PB^2-PD^2 = 0, \]and we are done. $\blacksquare$

May I be forgiven for my sins. First, if one of $\triangle PAB, PCD$ is equilateral, WLOG $\triangle PAB$, then well-known configuration properties get us $P, E_2, H_1=O_1$ are collinear, and the concurrency point is just the center of $\triangle PAB$. With this case, out of the way, assume neither $\triangle PAB$ nor $\triangle PCD$ is equilateral. We use moving points, twice. We say a point moves linearly if it moves along a fixed line at a constant speed. Lemma: Given non-equilateral $\triangle ABC$, let $X, Y$ be points on $AB, AC$ such that the orthocenter of $\triangle AXY$ lies on the Euler line of $\triangle ABC$. Then the line joining the nine-point centers $N_1, N_2$ of $\triangle ABC, \triangle AXY$ is perpendicular to $BC$. Proof. Let $O',H'$ be the circumcenter and orthocenter of $\triangle AXY$. Animate $H'$ linearly along the Euler line. Then $X, Y$ move linearly on $AB, AC$. Also, $\angle A$ is fixed, so $\triangle O'XY$ is similar to a fixed triangle and $O'$ moves linearly as well (e.g. if we apply complex numbers, then $O'$ is a linear function of $X, Y$ which themselves are linear in time). Now, $N_2=(O'+H')/2$ moves linearly, so it suffices to check that this line is the altitude from $N_1$ to $BC$ for two choices of $H'$. When $H'=H$, this is clear; when $H'$ is $O$, the circumcenter of $\triangle ABC$, then the circumcenter of $\triangle AXY$ lies on $AH$ and thus $N_1N_2 \parallel AH$. $\Box$ Now, let us return to the notations given in the problem. Let $\ell_1$ denote the perpendicular from $E_1$ to $CD$ and let $\ell_2$ denote the perpendicular from $E_2$ to $AB$. We'll fix $\triangle PAB$ and animate $C, D$ linearly on $PB, PD$ so that line $CD$ has fixed direction. Essentially, this is a homothety at $P$ in motion, so $H_2, O_2, E_2$ move linearly, while $H_1, \ell_1$ remain fixed. Moreover, if we define $X=\ell_1 \cap \ell_2$, then $E_2$'s motion implies that $X$ moves linearly as well. We wish to show that $H_1, X, H_2$ are collinear; it is easy to see (say, by shoelace) that this collinearity condition is equivalent to a quadratic in time being identically $0$. So, it suffices to check this for three choices of $CD$. When $CD$ passes through $P$, then $X=H_2$. When $CD$ is such that $H_2$ lies on $O_1H_1$, then we apply the above lemma on $\triangle PAB$. When $CD$ is such that $H_1$ lies on $O_2H_2$, then we apply the above lemma on $\triangle PCD$. Having checked these, we are done.

Here is a solution that mostly uses angle chasing and similar triangles, no length calculation / complex numbers / vectors / moving points / etc. Consequently, it is a bit involved. First, we prove the following important special case. Claim: Let $ABC$ be a triangle and $P$ a point on side $BC$ or its extension. Let $E_1$ and $E_2$ denote the nine-point centers of $\triangle ABP$ and $\triangle ACP$ respectively. The line through $E_1$ perpendicular to $\overline{AC}$ meets $\overline{AB}$ at $Z_1$; define $Z_2$ similarly. Then \[ \frac{AZ_1}{BZ_1} = \frac{AZ_2}{CZ_2} \]in the directed sense (i.e.\ $Z_1 Z_2 \parallel BC$). Proof. Let $X = \overline{E_1 Z_1} \cap \overline{E_2 Z_2}$. It suffices to show $X$ lies on the $A$-altitude (by homothety at $A$). Let $K$ be the altitude form $A$ and $M$, $N$ the midpoints of $\overline{AB}$ and $\overline{AC}$. Then $\triangle KE_1E_2 \sim \triangle KMN$ by Salmon theorem, as $E_1$ and $E_2$ are the circumcenters of $\triangle QMK$ and $\triangle QNK$. [asy][asy] pair A = (0,1.2); pair B = (-1,-1.4); pair C = (2.6,-1.4); pair K = foot(A, B, C); pair P = (0.58,-1.4); pair M = midpoint(A--B); pair N = midpoint(A--C); pair Q = extension(A, P, M, N); pair E_1 = circumcenter(M, Q, K); pair E_2 = circumcenter(Q, K, N); pair F_1 = foot(E_1, A, C); pair F_2 = foot(E_2, A, B); pair Z_1 = extension(E_1, F_1, A, B); pair Z_2 = extension(E_2, F_2, A, C); filldraw(A--B--C--cycle, invisible, red); filldraw(K--M--N--cycle, invisible, red); draw(Z_1--F_1, grey); draw(Z_2--F_2, grey); pair X = extension(E_1, F_1, E_2, F_2); draw(A--K, lightred+dotted); filldraw(K--E_1--E_2--cycle, invisible, deepcyan); draw(K--Q, deepcyan); draw(A--P, orange); filldraw(circumcircle(X, E_1, E_2), invisible, blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$K$", K, dir(K)); dot("$P$", P, dir(P)); dot("$M$", M, dir(M)); dot("$N$", N, dir(N)); dot("$Q$", Q, dir(80)); dot("$E_1$", E_1, dir(160)); dot("$E_2$", E_2, dir(20)); dot("$Z_1$", Z_1, dir(Z_1)); dot("$Z_2$", Z_2, dir(Z_2)); dot("$X$", X, dir(90)); /* TSQ Source: A = (0,1.2) B = (-1,-1.4) C = (2.6,-1.4) K = foot A B C P = (0.58,-1.4) M = midpoint A--B N = midpoint A--C Q = extension A P M N R80 E_1 = circumcenter M Q K R160 E_2 = circumcenter Q K N R20 F_1 := foot E_1 A C F_2 := foot E_2 A B Z_1 = extension E_1 F_1 A B Z_2 = extension E_2 F_2 A C A--B--C--cycle 0.1 lightred / red K--M--N--cycle 0.1 lightred / red Z_1--F_1 grey Z_2--F_2 grey X = extension E_1 F_1 E_2 F_2 R90 A--K lightred dotted K--E_1--E_2--cycle 0.1 lightcyan / deepcyan K--Q deepcyan A--P orange circumcircle X E_1 E_2 0.1 yellow / blue */ [/asy][/asy] First I claim $E_1 X E_2 K$ is cyclic. This is clear since $\measuredangle E_1 X E_2 = -\measuredangle BAC = \measuredangle E_1 K E_2$. Now, \[ \measuredangle X K E_1 = \measuredangle X E_2 E_1 = \measuredangle (\overline{AB}, \overline{QK}) \]since $\overline{XE_2} \perp \overline{AB}$, $\overline{E_1 E_2} \perp \overline{QK}$. On the other hand, $\measuredangle E_1KM = 90^{\circ} - \measuredangle MQK$. Thus $\measuredangle XKM = 90^{\circ} + \measuredangle (\overline{AB}, \overline{MQ}) = 90^{\circ} - \measuredangle MKB$ which proves $\measuredangle BKX = 90^{\circ}$. $\blacksquare$ Back to the main problem. We let the perpendicular from $E_1$ to $\overline{CD}$ meet $\overline{AB}$ at $Z_1$ and define $Z_2$ analogously. Claim: As in the previous claim, we have $\frac{AZ_1}{BZ_1} = \frac{DZ_2}{CZ_2}$. Proof. We gradually move line $CD$ continuously to the right in the figure above, until $D$ and $A$ coincide. This is equivalent to applying a homothety at $P$ affecting only the points $C$, $D$, $Z_2$, $E_2$. When $D=A$ we get the previous lemma. $\blacksquare$ [asy][asy] pair D = (-0.5862602838136073,1.9448319467989137); pair C = (0.2,0.6); pair B = (-1.2,0.6); pair A = (-0.8950982404878378,1.660701026658622); pair P = (-2.0480341390298196,0.6); pair E_1 = (-1.259557655001374,0.7823215176667854); pair E_2 = (-0.7551386766642586,1.1498656725186438); pair Q = (-0.7385490144461995,2.20530898513895); pair H_3 = (-0.46750431295872774,1.9106951791743063); pair H_4 = (-0.7385490144461996,0.8697887893957434); pair Z_1 = (-1.124972731991288,0.861007021840613); pair Z_2 = (0.006524694601061415,0.930923203134664); filldraw(A--B--C--D--cycle, invisible, red); draw(B--P--A--Q--D, red); pair X = extension(E_1, Z_1, E_2, Z_2); pair F_1 = foot(E_1, D, C); pair F_2 = foot(E_2, A, B); draw(E_1--F_1, deepgreen); draw(F_2--Z_2, deepgreen); draw(H_3--H_4, lightcyan); draw(D--H_3--A, lightgreen); draw(B--H_4--C, lightgreen); dot("$D$", D, dir(40)); dot("$C$", C, dir(270)); dot("$B$", B, dir(270)); dot("$A$", A, dir(A)); dot("$P$", P, dir(P)); dot("$E_1$", E_1, dir(E_1)); dot("$E_2$", E_2, dir(100)); dot("$Q$", Q, dir(Q)); dot("$H_3$", H_3, dir(20)); dot("$H_4$", H_4, dir(H_4)); dot("$Z_1$", Z_1, dir(Z_1)); dot("$Z_2$", Z_2, dir(10)); dot("$X$", X, dir(300)); /* TSQ Source: .D = (-0.5862602838136073,1.9448319467989137) R40 .C = (0.2,0.6) R270 .B = (-1.2,0.6) R270 A = (-0.8950982404878378,1.660701026658622) P = (-2.0480341390298196,0.6) E_1 = (-1.259557655001374,0.7823215176667854) .E_2 = (-0.7551386766642586,1.1498656725186438) R100 Q = (-0.7385490144461995,2.20530898513895) .H_3 = (-0.46750431295872774,1.9106951791743063) R20 H_4 = (-0.7385490144461996,0.8697887893957434) Z_1 = (-1.124972731991288,0.861007021840613) .Z_2 = (0.006524694601061415,0.930923203134664) R10 A--B--C--D--cycle 0.1 lightred / red B--P--A--Q--D red X = extension E_1 Z_1 E_2 Z_2 R300 F_1 := foot E_1 D C F_2 := foot E_2 A B E_1--F_1 deepgreen F_2--Z_2 deepgreen H_3--H_4 lightcyan D--H_3--A lightgreen B--H_4--C lightgreen */ [/asy][/asy] Finally, we let $H_3$ and $H_4$ denote the orthocenters of $\triangle QAD$ and $\triangle QBC$. By the Gauss-Bodenmiller theorem, line $H_3 H_4$ coincides with line $H_1 H_2$. We define $X$ on $\overline{H_3 H_4}$ such that \[ \frac{AZ_1}{BZ_1} = \frac{H_3X}{H_4X} = \frac{DZ_2}{CZ_2} \]It follows that $\overline{XZ_1} \perp \overline{CD}$ and $\overline{XZ_2} \perp \overline{AB}$ from the ratios, and thus $X$ is the desired concurrency point. Remark: I believe in the process of solving the problem, one naturally discovers all three claims in reverse order: first eliminating $H_1$ and $H_2$ using Gauss Bodenmiller, then realizing that the statement involves only angles (i.e.\ that we can move $\overline{CD}$ until $D=A$), and then solving synthetically the reduced problem.

To amend for my controversial trolling today, here is a legitimate post which showcases a possibly new solution:

wait this post is also a troll half of this solution is extraneous T_T

Solved with Twistya, v4913. Claim: $\triangle PO_1O_2 \sim \triangle PH_1H_2.$ Proof: $\angle O_1PO_2 = \angle H_1PH_2$ is simple. Note $PH_1 = \frac{2PO_1}{\cos(\angle APB)}$ and $PH_2 = \frac{2PO_2}{\cos(\angle APB)}$ so $\frac{PH_1}{PH_2}= \frac{PO_1}{PO_2}.$ $\square$ Let $H_2O_2,H_1O_1$ meet $PH_1,PH_2$ at $K_2,K_1$ respectively. To prove that the perpendicular from $E_1$ on $CD$ and the perpendicular from $E_2$ on $AB$ intersect $H_1H_2$ at the same point, it suffices to show $\frac{H_2E_2}{E_2K_2} = \frac{H_1E_1}{E_1K_1}.$ This holds if $\frac{H_2K_2}{O_2K_2} = \frac{H_1K_1}{O_1K_1}.$ By Law of Sines, \begin{align*} \frac{O_1K_1}{O_2K_2} &= \frac{O_1P/\sin(\angle O_1K_1P)}{O_2P/\sin(\angle O_2K_2P)}\\ &= \frac{H_1P/\sin(\angle H_1K_1P)}{H_2P/\sin(\angle H_2K_2P)}\\ &= \frac{H_1K_1}{H_2K_2} \end{align*}so we're done. $\blacksquare$

[asy][asy] //09SLG6 //setup; size(7cm); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter //defn pair P,A,D,B,C; P=(0,0); A=(26,0); D=(5,0); B=(19,14); C=.6*B; pair O1,H1,E1,O2,H2,E2; O1=circumcenter(P,A,B); H1=orthocenter(P,A,B); E1=(O1+H1)/2; O2=circumcenter(P,C,D); H2=orthocenter(P,C,D); E2=(O2+H2)/2; pair X=extension(E1,foot(E1,C,D),E2,foot(E2,A,B)); //draw fill(A--B--C--D--cycle,blu1); fill(P--H1--H2--cycle,RGB(255,230,255)); fill(D--extension(H1,H2,A,D)--H1--extension(P,H1,C,D)--cycle,lightpurple); draw(C--D--P--B--A--D,blu); draw(P--H1--H2--P,magenta); draw(3*X-2*E1--3.5*E1-2.5*X ^^ 1.7*E2-.7*X--1.3*X-.3*E2 ,purple); draw(O1--H1^^O2--H2,red); draw(H2--O1--P--O2--H1,red+dotted); //label void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.17),a,linewidth(.3)); label(s,P,v);} string labels[]={"$P$","$A$","$B$","$C$","$D$", "$O_1$","$H_1$","$E_1$", "$O_2$","$H_2$","$E_2$",""}; //12 pair points[]={P,A,B,C,D,O1,H1,E1,O2,H2,E2,X}; real dirs[]={200,40,90,220,120, -60,110,90, 110,-90,-100, -80}; pen colors[]={blu,blu,blu,blu,blu, red,magenta,purple, red,magenta,purple, magenta}; for (int i=0; i<12; ++i) {pt(labels[i],points[i],dir(dirs[i]),colors[i]); } [/asy][/asy] Trying not to bash excessively\dots consider the problem wrt $\triangle PH_1H_2$. Observe that by isogonals, \[\frac{PH_1}{PO_1}=2\cos P=\frac{PH_2}{PO_2}\Rightarrow [PO_2H_1]=[PO_1H_2] \Rightarrow h_1(O_1)=-h_2(O_2) \overset{\text{linearity}}\Rightarrow \boxed{h_1(E_1)+h_2(E_2)=1}\]in barycentrics wrt $\triangle PH_1H_2$, where $p(X)$ denotes the $P$-coordinate of $X$, and similarly for the $H_k$. This means that the three desired lines (which can be defined as those through $E_1,E_2$ parallel to $\overline{PH_2}$, $\overline{PH_1}$ respectively) concur at \[\boxed{0P+h_1(E_1)\cdot H_1+h_2(E_2)\cdot H_2}\in\overline{H_1H_2}\]which is a valid barycentric point because of the first boxed equation.

I don't think this approach has been posted above as far as I saw.

), so we want to show that $E$ also lies on the radical axis of these circles $\omega_{AC}$ and $\omega_{BD}$. To do that, define the function $f:\mathbb{R}^2 \to \mathbb{R}$ \[f(X) = \mathbb{P}(X, \omega_{BD}) - \mathbb{P}(X,\omega_{AC})\]Here comes the tricky part: take a leap of faith and introduce $H'$ to be the intersection of the perpendiculars from $H_{1}$ and $H_{2}$ to $\overline{CD}$ and $\overline{AB}$, respectively, and $O'$ to be the intersection of the perpendiculars from $O_{1}$ and $O_{2}$ to $\overline{CD}$ and $\overline{AB}$, respectively. Notice that $E$ is the midpoint of $O'H'$ because of the two triplets of parallel lines $O_{1}O'\parallel E_{1}E \parallel H_{1}H'$ and $O_{2}O'\parallel E_{2}E \parallel H_{2}H'$. Hence, by Linearity of PoP, we have that: \[f(E) = \frac{1}{2}f(O')+\frac{1}{2}f(H')\]However, $H'$ is defined in such a way that $PH_{1}H'H_{2}$ is a parallelogram and $H_{1}H_{2}$ is the radical axis of $\omega_{AC}$ and $\omega_{BD}$, so $f(H') = -f(P)$. Thus we've reduced the problem to just showing that $PO'\parallel H_{1}H_{2}$ as this would imply that $f(O') = f(P)\Longrightarrow f(E) = 0$. We'll now prove two claims that finish the problem: Claim 1. $P, O', O_{1}, O_{2}$ are concyclic. Proof: Let $\overline{O'O_{1}} \cap \overline{CD} = X$, $\overline{O'O_{2}} \cap \overline{AB} = Y$ and $Q = \overline{AB} \cap \overline{CD}$. Because $O'O_{1} \perp CD$ and $O'O_{2} \cap AB$, so $O'XYQ$ is cyclic. Therefore \[\angle O_{1}PO_{2} = \angle XQY = \angle H_{1}PH_{2} = \angle O_{1}PO_{2}\]so $PO'O_{1}O_{2}$ is cyclic. Claim 2. $\triangle PO_{1}O_{2} \sim \triangle PH_{1}H_{2}$ Proof: The pairs $(O_{i}, H_{i})$ are isogonal wrt $\angle APB$, so $\angle O_{1}PO_{2} = \angle H_{1}PH_{2}$. Also, from $\triangle ABP$ and $\triangle CDP$ we have that: \[\frac{PH_{1}}{PO_{1}} = 2\cos\angle APB = 2\cos\angle CPD = \frac{PH_{2}}{PO_{2}}\]so we also have $\frac{PO_{1}}{PO_{2}} = \frac{PH_{1}}{PH_{2}}$, proving the desired similarity. Finally, by combining the two claims, we get: \[PO'O_{2} = \angle PO_{1}O_{2} = \angle PH_{1}H_{2} = \angle(\overline{O'O_{2}}, \overline{H_{1}H_{2}})\]and as $\angle(\overline{O'P}, \overline{O'O_{2}}) = \angle PO'O_{2} = \angle (\overline{O'O_{2}}, \overline{H_{1}H_{2}})$ we get $PO'\parallel H_{1}H_{2}$, so we're done.

The following solution is similar to the above post, but has a completely computational ending. Let $E$ denote the intersection of the perpendicular from $E_1$ to $\overline{CD}$ and $E_2$ to $\overline{AB}$. Claim: $E$ is the midpoint of the segment whose endpoints are the intersection of the perpendicular from $O_1$ to $\overline{CD}$ and $O_2$ to $\overline{AB}$ (denote by $O$), and the intersection of the perpendicular from $H_1$ to $\overline{CD}$ and $H_2$ to $\overline{AB}$ (denote by $H$). Proof. Trivial by projection chasing. Let $f(\bullet) = \text{Pow}(\bullet, (AC)) - \text{Pow}(\bullet, (BD))$ for each $\bullet$ in $\mathbb{R}^2$. By the Gauss-Bodenmiller theorem, $\overline{H_1H_2}$ is the radical axis of $(AC)$ and $(BD)$. Claim: We have $f(H)=-f(P)$. Proof. I contend that in the vectorial sense, $H=H_1+H_2-P$. It suffices to show that $H'=H_1+H_2-P$ satisfies $(H'-H_1) \cdot (C-D) = 0$ and $(H'-H_2) \cdot (A-B) = 0$. However, plugging in the value of $H'$ returns $(H_2-P) \cdot (C-D) = 0$ and $(H_1-P) \cdot (A-B) = 0$, respectively, which are evidently true. Thus, \[ f(H) = f(H_1+H_2-P) = f(H_1)+f(H_2)-f(P) = 0+0-f(P) = -f(P), \]as claimed. Combining the preceding two claims, note that \[ f(E) = f((O+H)/2) = (f(O)+f(H))/2 = (f(O)-f(P))/2, \]so if suffices to show that $f(O)=f(P)$. Let $O'$ denote the intersection of the perpendicular bisectors of $\overline{AB}$ and $\overline{CD}$. It is not hard to see that $O'O_1OO_2$ is a parallelogram, so \[ f(O) = f(O_1)+f(O_2)-f(O'). \] Claim: Explicitly, we have \[ f(O') = \frac{BD^2-AC^2}{2}. \]Proof. We use vectors. Note that \[ (O'-(A+B)/2) \cdot (A-B) = 0 \]and \[ (O'-(C+D)/2) \cdot (C-D) = 0. \]Summing these equations, we obtain \[ (B+D+A-C) \cdot O' = (B^2+D^2-A^2-C^2)/2. \]Meanwhile, note that the desired quantity is \[ (O'-(A+C)/2)^2 + (O'-(B+D)/2)^2 - (AC^2-BD^2)/4. \]The expression $(O'-(A+C)/2)^2 + (O'-(B+D)/2)^2$ rewrites as \[ (B+D+A-C) \cdot O' - (B+D-A-C)/2 \cdot (A+B+C+D)/2 = \frac{(B-D)^2-(A-C)^2}{4} = \frac{BD^2-AC^2}{4}. \]Thus \[ f(O') = \frac{BD^2-AC^2}{2}, \]as desired. Realize that \[ P+A+B-2O_1=H_1 \]and \[ P+C+D-2O_2=H_2, \]so summing these two equations we have \[ 2P+(A+B+C+D)-2(O_1+O_2)=H_1+H_2. \]Taking $f$ of both sides and leveraging linearity, we compute \[ 2f(P)+(f(A)+f(B)+f(C)+f(D))-2(f(O_1)+f(O_2))=f(H_1)+f(H_2)=0+0=0. \]We will solve for $f(O_1)+f(O_2)$. A quick Apollonius calculation computes $f(\bullet)$ for $\bullet=A, B, C, D$. Plugging in these values returns \[ f(O_1)+f(O_2) = f(P) + \frac{BD^2-AC^2}{2} = f(P)+f(O'), \]which implies $f(O)=f(P)$, as desired.

The perpendicular from $E_2$ to $AB$ is parallel to $CD$. Let it intersect $H_1H_2$ at $P_2$ and let $O_2H_2$ intersect $PH_1$ at $Q_2$. \[\frac{P_2H_2}{H_1H_2}=\frac{E_2H_2}{Q_2H_2}\]Define $P_1,Q_1$ analogously, then we have \[\frac{P_1H_1}{H_1H_2}=\frac{E_1H_1}{Q_1H_1}\]and it just suffices to show that the sum of the two numbers above is $1$. Since $E_1$ and $E_2$ are midpoints of $O_1H_1$ and $O_2H_2$, it suffices to show \[\frac{O_2H_2}{Q_2H_2}+\frac{O_1H_1}{Q_1H_1}=2\]Note that \[\frac{O_1Q_1}{O_1P}=\frac{\sin(\angle O_1PH_2)}{\sin(\angle PQ_1O_1)}\]\[\frac{O_2Q_2}{O_2P}=\frac{\sin(\angle O_2PH_1)}{\sin(\angle PQ_2O_2)}\]Since $PH_1$ and $PO_1$ are isogonal, and $PH_2$ and $PO_2$ are isogonal, we have $\angle O_1PH_2=\angle O_2PH_1$ which implies that \[\frac{O_1Q_1}{O_2Q_2}=\frac{O_1P \sin(\angle PQ_2O_2)}{O_2P \sin(\angle PQ_1O_1)}\]Similarly, \[\frac{H_1Q_1}{H_2Q_2}=\frac{H_1P \sin(\angle PQ_2H_2)}{H_2P \sin(\angle PQ_1H_1)}\]Note that all the angles are the same. Since $\tfrac{O_1P}{H_1P}$ depends only on the value $\angle P$, the two expressions are equal. The result follows.

Me when reflecting one way does nothing but reflecting the other way wins. Okay, whatever. One can compute $\triangle PO_1O_2 \sim \triangle PH_1H_2$. Reflect $P$ over $H_1H-2$ to $P'$ and let $P''$ be the midpoint of $PP'$. Let $(P''E_1E_2)$ meet $P''H_1H_2$ again at $X$. We have \[ \measuredangle E_1XP'' = \measuredangle E_1E_2P'' = \measuredangle PH_2H_1 = \measuredangle PH_2P \]implying $XE_1 \parallel PH_2 \perp CD$, and similarly so for the other one. Buh.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(16cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.81822432833631, xmax = 18.53621426953341, ymin = -1.0521366196519075, ymax = 15.667814580424341; /* image dimensions */ /* draw figures */ draw((-0.6292668998753987,13.84285103014296)--(-4.434730693901156,1.2536288880224444), linewidth(1.)); draw((-0.6292668998753987,13.84285103014296)--(8.96,0.66), linewidth(1.)); draw((xmin, 1.9542979961519296*xmin + 5.803313140713537)--(xmax, 1.9542979961519296*xmax + 5.803313140713537), linewidth(1.)); /* line */ draw((-1.4791694866379892,9.820021635457167)--(1.9389629318155697,9.592624512973575), linewidth(1.)); draw((2.4875104398435726,6.030945208976445)--(-1.0790184734636994,3.694589500212516), linewidth(1.)); draw((-0.6292668998753987,13.84285103014296)--(-1.0790184734636994,3.694589500212516), linewidth(1.)); draw((-0.6292668998753987,13.84285103014296)--(1.9389629318155697,9.592624512973575), linewidth(1.)); draw((-0.6292668998753987,13.84285103014296)--(-1.4791694866379892,9.820021635457167), linewidth(1.)); draw((-0.6292668998753987,13.84285103014296)--(2.4875104398435726,6.030945208976445), linewidth(1.)); draw((2.4875104398435726,6.030945208976445)--(-1.4791694866379892,9.820021635457167), linewidth(1.)); draw((-4.434730693901156,1.2536288880224444)--(8.96,0.66), linewidth(1.)); draw((-3.,6.)--(2.6098908584149902,9.38981675374495), linewidth(1.)); /* dots and labels */ dot((-3.,6.),dotstyle); label("$A$", (-3.4587101451455364,5.955770033683012), W * labelscalefactor); dot((-4.434730693901156,1.2536288880224444),dotstyle); label("$D$", (-4.677476519638569,0.6998400436818223), S * labelscalefactor); dot((8.96,0.66),dotstyle); label("$C$", (9.033645193408045,0.852185840493451), SE * labelscalefactor); dot((2.6098908584149902,9.38981675374495),dotstyle); label("$B$", (2.6922514011239866,9.573982707959193), NE * labelscalefactor); dot((-0.6292668998753987,13.84285103014296),linewidth(4.pt) + dotstyle); label("$P$", (-0.5450967811231308,13.992010815496425), NE * labelscalefactor); dot((-1.0790184734636994,3.694589500212516),linewidth(4.pt) + dotstyle); label("$H_2$", (-1.0211773961594717,3.365891487885324), SE * labelscalefactor); dot((1.9389629318155697,9.592624512973575),linewidth(4.pt) + dotstyle); label("$H_1$", (2.0828682138774703,9.478766584951925), S * labelscalefactor); dot((2.4875104398435726,6.030945208976445),linewidth(4.pt) + dotstyle); label("$O_2$", (2.558948828913811,6.184288728900455), NE * labelscalefactor); dot((-1.4791694866379892,9.820021635457167),linewidth(4.pt) + dotstyle); label("$O_1$", (-1.7448199310147094,9.364507237343204), S * labelscalefactor); dot((-1.8143221497666757,2.2575869990504627),linewidth(4.pt) + dotstyle); label("$E$", (-1.6496038080074413,1.956692867377759), NE * labelscalefactor); dot((1.2036592555125947,8.155622011811525),linewidth(4.pt) + dotstyle); label("$F$", (1.2830527806164178,7.726789921618195), NE * labelscalefactor); dot((-0.19505457079250488,7.694908376872475),linewidth(4.pt) + dotstyle); label("$M$", (-0.1261458398911509,7.841049269226917), S * labelscalefactor); dot((2.2626346530494224,0.9568144440112223),linewidth(4.pt) + dotstyle); label("$N$", (2.3304301336963675,1.0997477603123476), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Denote by $M$ and $N$ the midpoints of $AB$ and $CD$ respectively. It's well known that $PH_1 = 2O_1M$ and $PH_2 = 2O_2N$ so $$\frac{PH_1}{PH_2} = \frac{O_1M}{O_2N} = \frac{PO_1}{PO_2}$$Combining this with the fact that $\angle H_2PH_1 = \angle O_1PO_2$ we deduce that $\triangle PO_1O_2 \sim \triangle PH_1H_2$. Note that these triangles are inversely oriented. Next, let the perpendicular from $O_1$ to $CD$, and the perpendicular from $O_2$ to $AB$ intersect $H_1H_2$ at $E$ and $F$ respectively. Notice that the perpendicular from $E_1$ to $CD$ intersects $H_1H_2$ at the midpoint of $H_1E$ and similarly the perpendicular from $H_2$ to $AB$ intersects $H_1H_2$ at the midpoint of $H_2F$. Thus, it suffices to show that the aforementioned midpoints coincide, which is equivalent to showing that $EF = H_1H_2$. Using the fact that $O_1E \parallel PH_2$ and $O_2F \parallel PH_1$ we may restrict our attention to the triangles $\triangle PO_1O_2$ and $\triangle PH_1H_2$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.1506711393152, xmax = 12.666605450307435, ymin = 1.613217979818628, ymax = 14.438155499352822; /* image dimensions */ pen wwccff = rgb(0.4,0.8,1.); pen ccccff = rgb(0.8,0.8,1.); draw((-0.6292668998753987,13.84285103014296)--(-1.0790184734636994,3.694589500212516)--(1.9389629318155697,9.592624512973575)--cycle, linewidth(1.) + wwccff); draw((-0.6292668998753987,13.84285103014296)--(-1.4791694866379892,9.820021635457167)--(2.4875104398435726,6.030945208976445)--cycle, linewidth(1.) + ccccff); /* draw figures */ draw((-1.4791694866379892,9.820021635457167)--(1.9389629318155697,9.592624512973575), linewidth(1.)); draw((2.4875104398435726,6.030945208976445)--(-1.0790184734636994,3.694589500212516), linewidth(1.)); draw((-0.6292668998753987,13.84285103014296)--(-1.4791694866379892,9.820021635457167), linewidth(1.)); draw((-0.6292668998753987,13.84285103014296)--(2.4875104398435726,6.030945208976445), linewidth(1.)); draw((2.4875104398435726,6.030945208976445)--(-1.4791694866379892,9.820021635457167), linewidth(1.)); draw((1.9389629318155697,9.592624512973575)--(-1.8143221497666757,2.2575869990504627), linewidth(1.)); draw((-1.3645705761783744,12.405848528980913)--(-1.8143221497666757,2.2575869990504627), linewidth(1.) + wwccff); draw((-1.3645705761783744,12.405848528980913)--(2.4875104398435726,6.030945208976445), linewidth(1.) + wwccff); draw((-0.6292668998753987,13.84285103014296)--(-1.0790184734636994,3.694589500212516), linewidth(1.) + wwccff); draw((-1.0790184734636994,3.694589500212516)--(1.9389629318155697,9.592624512973575), linewidth(1.) + wwccff); draw((1.9389629318155697,9.592624512973575)--(-0.6292668998753987,13.84285103014296), linewidth(1.) + wwccff); draw((-0.6292668998753987,13.84285103014296)--(-1.4791694866379892,9.820021635457167), linewidth(1.) + ccccff); draw((-1.4791694866379892,9.820021635457167)--(2.4875104398435726,6.030945208976445), linewidth(1.) + ccccff); draw((2.4875104398435726,6.030945208976445)--(-0.6292668998753987,13.84285103014296), linewidth(1.) + ccccff); draw(circle((3.3470223466070133,10.901586877902483), 4.9458984001843955), linewidth(1.)); /* dots and labels */ dot((-0.6292668998753987,13.84285103014296),linewidth(4.pt) + dotstyle); label("$P$", (-0.8740745959434084,14.02915976980048), NE * labelscalefactor); dot((-1.0790184734636994,3.694589500212516),linewidth(4.pt) + dotstyle); label("$H_2$", (-0.9909305186726498,3.351449830416087), NE * labelscalefactor); dot((1.9389629318155697,9.592624512973575),linewidth(4.pt) + dotstyle); label("$H_1$", (2.061930462628781,9.500992764042392), NE * labelscalefactor); dot((2.4875104398435726,6.030945208976445),linewidth(4.pt) + dotstyle); label("$O_2$", (2.5439611438869014,6.141384985576715), NE * labelscalefactor); dot((-1.4791694866379892,9.820021635457167),linewidth(4.pt) + dotstyle); label("$O_1$", (-1.9257779005065807,9.720097619159718), NE * labelscalefactor); dot((-1.8143221497666757,2.2575869990504627),linewidth(4.pt) + dotstyle); label("$E$", (-1.6774590647069427,2.022213709370971), NE * labelscalefactor); dot((1.2036592555125947,8.155622011811525),linewidth(4.pt) + dotstyle); label("$F$", (0.8495502643129017,8.06950771060919), NE * labelscalefactor); dot((-1.3645705761783744,12.405848528980913),linewidth(4.pt) + dotstyle); label("$X$", (-1.940384890847736,12.510032774320345), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $X = O_1E \cap O_2F$. Then we have $\angle O_1XO_2 = \angle EXF = \angle H_2PH_1 = \angle O_1PO_2$ so we get that $P$, $X$, $O_1$ and $O_2$ are cyclic. Next note that $\angle EXP = \angle O_1XP = \angle O_1O_2P = \angle PH_2H_1 = \angle XEH_2$ so we have $XP \parallel EH_2$ i.e $PXEH_2$ is a parallelogram. Thus we get $XE = PH_2$. Similarly, we also have $XF = PH_1$, from which we deduce that $\triangle XEF \equiv \triangle PH_2H_1$ so $EF = H_2H_1$ as desired.