$\angle APX+\angle AQX=360^{\circ}-\angle ADP-\angle ADQ=180^{\circ}$ so $APQX$ is cyclic. Since $\angle PAD=\angle DPX=\angle QAX$, we get $\triangle PAD\sim\triangle XAQ$. Now, we will use complex numbers. Let $A$ lie on the unit circle. Then, $\frac{d-a}{d-p}=\frac{q-a}{q-x}$, so $q-x=\frac{(q-a)(d-p)}{d-a}=y+\frac z{d-a}$ for some constants $y$ and $z$. Since $d$ lies on the circle, $\frac z{d-a}$ is a line by inversion, so $X$ lies on a line.

Sol:- Invert wrt a circle centered at $D$ with arbitrary radius , we get the following problem. Inverted Problem Let $D$ be a point on segment $PQ$. Let $l$ be a fixed line not passing through $D$ and let $A$ be a variable point on $l$.Let $X$ be the intersection of the circle passing through $D,P$ tangent to $AP$ and the circle passing through $D,Q$ tangent to $AQ$.Show that as $A$ varies,$X$ lies on a fixed circle. We label the perpendicular bisector of $PQ$ as line $m$.Let $A'$ be the reflection of $A$ across $m$.Reflection of line $l$ across $m$ be $l'$. As $A$ varies on $l$, $A'$ varies on $l'$. Step 1 $APXQA'$ is cyclic and $X,D,A'$ are collinear. ProofBy symmetry clearly $A' \in (APQ)$.$\measuredangle PXQ=\measuredangle PXD+\measuredangle DXQ=\measuredangle APQ+\measuredangle PQA=\measuredangle PAQ$. So $X \in (APQ)$.$\measuredangle PXD=\measuredangle APQ=\measuredangle PQA'=\measuredangle PXA'$. So $X,D,A'$ are collinear. Let the concurrence point of $l,l',m$ be $J$.Clearly $J$ is a fixed point.$JD$ meet $(JPQ)$ again at $K$.So $K$ is also a fixed point. Step 2 $X$ lies on a fixed circle. Proof Since $D \in PQ$, $D$ has equal power wrt $(JPQ)$ and $(A'PQ)$ so $XKA'J$ is cyclic.Hence $\measuredangle DJA'=\measuredangle DXK$ and since $\measuredangle DJA'$ is a fixed angle so $\measuredangle DXK$ is a fixed angle subtended on fixed segment $DK$ implying that $(DXK)$ is a fixed circle.

Steiner conic! We will prove, in turn, that i) There is a projective map from $B(P)$, the pencils of lines through $P$, to $B(Q)$ the pencil of lines through $Q$ given by $PX\to QX$ and ii) This map is a perspecitvity, meaning that $X=PX\cap QX$ lies on a fixed line. Proof of i): Define $P'=AP\cap \omega\neq A$, $Q'=AQ\cap \omega\neq A$, and further $D'$ is the second intersection of $\omega$ with $PQ$ which can only coincide with $D$ is these are tangent; clearly (key synthetic observation!) $DP'\parallel PX$, $DQ'\parallel QX$. Now let $l$ be the ideal line. We have a projective mapping $$A\to QA\cap \omega=Q'\to D'Q'\to D'Q'\cap l\to QX$$and a symmetrical one proving that $A\to PX$ is projective as well. Hence $PX\to A\to QX$ is indeed projective. $\square$ Proof of ii) By a special case of the Steiner conic, it suffices to show that the image of $PQ$ in the above mapping is itself. We prove this in the limit, which suffices since any projective map is continuous. Take the limit $A\to D'$. Then $P'=Q'=D$, so $PX=PQ=QX$ in this case indeed. $\square$.

Nice problem! To begin, we will first prove a lemma. Lemma. Let $P$ be a fixed point and $P\in\tau$ be a fixed circle. Let $P_1\in\tau$ and $P_2$ be a point such that $PP_1\cdot PP_2$ is constant. Then, as $P_1$ varies, $P_2$ lies on a fixed line. Proof. We perform an inversion about a circle with center $P$ and radius $\sqrt{|PP_1\cdot PP_2|}$. Since $\tau$ passes through $P$, it becomes a line. Under this inversion (and reflection about $P$ if required), $P_1$ goes to $P_2$. So, $P_2$ lies on a line, as claimed. $\square$ $\noindent\rule{20.5cm}{0.1pt}$ Note that since $\angle PAD=\angle XPD=\angle XPQ$ and $\angle DAQ=\angle DQX=\angle PQX$, we have $$\angle PAQ=\angle PAD+\angle DAQ=\angle XPQ+\angle PQX=\angle PXQ\Longrightarrow X\in(APQ)$$Let $AD$ be extended to meet $(APQ)$ at $B$. Then, $\angle PAB=\angle XPQ=\angle XAQ\Longrightarrow PB=XQ\Longrightarrow PBXQ$ is an isosceles trapezoid. So, $PQ\parallel BX$. Note that $DA\cdot DB=DP\cdot DQ$ which is constant. So, from our Lemma, $B$ lies on a line. Since $B$ is the reflection of $X$ over the perpendicular bisector of $PQ$, we conclude that $X$ lies on a line, as desired. $\square$

Let's break the symmetry. Invert at $P$ with arbitrary radius. Then $A$ still lies on a fixed circle through $D$, and $D,P,Q$ are still collinear. The tangent at $P$ inverts to the line through $P$ parallel to $AD$, and the tangent at $Q$ inverts to the circle through $P,Q$ tangent to $(ADQ)$ at $Q$. Then the homothety through $Q$ taking $D$ to $P$ takes $(ADQ)$ to the tangent circle through $Q$ and takes $AD$ to the parallel line through $P$, so $A$ is taken to the intersection of these, which is $X$. Thus, $X$ is the image of $A$ under a fixed dilation, and so lies on a fixed circle through $P$. Inverting back means $X$ lies on a fixed line. $\square$

Most efficient synthetic solution? Let $\measuredangle$ denote directed angles $\!\!\!\mod{180^\circ}$. We have $\measuredangle APX=\measuredangle ADP=\measuredangle ADQ=\measuredangle AQX$, so $APXQ$ is cyclic. Let $\overline{AD}$ intersect the circumcircle of $APXQ$ again at $X'$. Notice that $X'$ is on the image of $\omega$ after an inversion centered at $D$ with power $-DP \cdot DQ$, which is a fixed line. Since $\measuredangle PAX'=\measuredangle XPQ=\measuredangle XAQ$, we know that $X$ is the reflection of $X'$ over the perpendicular bisector of $\overline{PQ}$, so $X$ lies on a fixed line. $\square$

All geometry is a lie. We use moving points. Claim: Let $\ell$ be a degree $1$ line and $P$ be a fixed point. Then the line through $P$ perpendicular to $\ell$ also has degree $1$. Proof: Let $A$ be the (unique) point that $\ell$ always passes through (exists since degree 1 implies fixed point), and animate another point $B$ on it linearly. Then do coordinate algebra, WLOG letting $A$ be the origin and having $B$ vary along the $x$-axis. In this case the reciprocal of the slope of $\ell$ varies linearly, so the slope of the perpendicular through $P$ varies linearly, which implies the desired result. Let $O$ be the center of $\omega$, $O_1$ be the center of $(ADP)$, and $O_2$ be the center of $(ADQ)$. Animate $A$ with degree $2$ on $\omega$, so $\overline{DA}$ has degree $1$. By the claim, the perpendicular bisector of $\overline{DA}$ (which passes through $O$) has degree $1$. Thus $O_1$, which is the intersection of this perpendicular bisector with the fixed perpendicular bisector of $\overline{PD}$, has degree $1$, as does $O_2$. Then $\overline{O_1P}$ (resp. $\overline{O_2Q}$) has degree $1$, so by the claim the tangent to $(ADP)$ at $P$ (resp. tangent to $(ADQ)$ at $Q$) has degree $1$ as well. Hence their intersection $X$ has degree at most $2$. But when $A$ lies on line $\overline{PQ}$, $O_1$ and $O_2$ are both at infinity, hence the "tangents" at $P$ and $Q$ coincide (both with $\overline{PQ}$), so by "Strong Zack's Lemma" $X$ has degree (at most) $1$, which finishes the problem. $\blacksquare$

Let $D'$ be the reflection of $D$ over the midpoint of $\overline{PQ}$. Claim: we have $\triangle PD'X \sim \triangle ADP$ and $\triangle QD'X \sim \triangle ADQ$. Proof: let $E$ be the unique point on $\overline{PQ}$ such that $\triangle PEX \sim \triangle ADP$. Then, since \[\angle QEX = 180^{\circ} - \angle PEX = \angle 180^{\circ} - \angle ADP = \angle ADQ,\]it follows that $\triangle QEX \sim \triangle ADQ$. Then, we have \[ \frac{PE}{EQ} = \frac{\frac{PE}{EX}}{\frac{QE}{EX}} = \frac{\frac{AD}{DP}}{\frac{AD}{DQ}} = \frac{QD}{DP}, \]so $E = D'$. As a consequence, $\angle XAP = \angle DAQ$. Also, it's clear from angle chasing that $APXQ$ is cyclic. If we extend line $AD$ to meet $(APXQ)$ again at $X'$, then $X'$ is the reflection of $X$ over the perpendicular bisector of $\overline{PQ}$. But clearly, $X'$ lies on a fixed line: the image of $\omega$ under an inversion at $D$ with radius $-PD \cdot DQ$.

Uh kinda cute ! . [this solution i did not account for config issues you can fix them yourself using directed angles] Here is my solve:- Let $\omega_1$ be the circumcircle of $\Delta ADP$ and $\omega_2$ that of $\Delta ADQ$ . Suppose $PA$ intersect $\omega_2$ again at $C$ , define $B$ analogously . Extend $PB$ and $CQ$ to meet at $E$ . Claim: $EQXP$ is a parallelogram . Proof: $\angle EQP=\angle PAD=\angle QPX$ so $EQ|| XP$ similarly $EP|| QX$. Hence $E$ is the reflectinon of $X$ about the midpoint of $PQ$ so it suffices to show $E$ moves on a line. say $\omega$ intersects $PQ$ at point $T \neq D$ and construct point $J$ on ray $PQ$ such that $\angle{TAD}=\angle{EJQ}$ Claim: $J$ stays fixed . Proof: Let $\omega$ intersect $AP$ at $F \neq A$ and $\omega$ intersect $AQ$ at $G \neq A$ Observe $(P,Q;T,D)=(F,G;T,D)$ [projecting through $A$] And $(F,G;T,D)=\frac{\sin(\angle{FAT})}{\sin(\angle{GAT})} / \frac{\sin(\angle{FAD})}{\sin(\angle{GAD})}$ Now just observe $\angle{FAD}=\angle{FAT}+\angle{TAD}$ and $\angle{GAD}=\angle{GAT}-\angle{TAD}$ . Finally observe $\angle{EQP}=\angle{PAD}$ and $\angle{EPQ}=\angle{DAQ}$ with bit of sine rule u get that $\frac{JQ}{JP}$ stays fixed hence $J$ also . by this rest is trivial

Invert the diagram at $D$. The problem becomes the following: Quote: Let $D$ be a point on segment $PQ$. Let $\ell$ be a fixed line and let $A$ be a variable point on $\ell$. Let $X$ be the intersection, other than $D$, of the circle through $D$ tangent to $AP$ at $P$ and the circle through $D$ tangent to $AQ$ at $Q$. Show that as $A$ varies, $X$ lies on a fixed circle containing $D$. Let $B$ be the intersection of $\ell$ with $(APQ)$, other than $A$. Let $F$ be the intersection of $\ell$ and $PQ$. Let $X'$ be the intersection of $(BDF)$ with $(APQ)$, other than $B$. Then $\angle APX' = \angle ABX' = 180^\circ - \angle PDX'$, so $(PDX')$ is tangent to $AP$ at $P$. Similarly $(PDX')$ is tangent to $AQ$ at $Q$, so $X = X'$. Let $C$ be the intersection of $BX$ with $PQ$. Then $CD \cdot CF = CB \cdot CX = CP \cdot CQ$, so $C$ can be defined in terms of $D$, $F$, $P$, and $Q$ alone, and $C$ is independent of $A$. Consider the inversion centered at $C$ swapping $D$ and $F$; this also swaps $B$ and $X$. This inversion is also independent of $A$. The image of $\ell$ under this inversion is a fixed circle containing $D$ and $X$ as desired.

$\textbf{Claim I:}$ $(APXQ)$ \[\measuredangle PAQ= \measuredangle PAD+\measuredangle QAD=\measuredangle PQX +\measuredangle PQX=\measuredangle PXQ\]This gives $AX$ and $AD$ to be isogonal since $\angle XPQ=\angle PAD$ and $\angle PQX=\angle DAQ.$ Let $\overline{AD}\cap (APXQ)=X'$. Notice that $X'$ is on the image of $\omega$ after a negative inversion with center $D$ and radius $\sqrt{DP \cdot DQ}$. Also note that $X$ is the reflection of $X'$ over the perpendicular bisector of $\overline{PQ}$. Because $X’$ lies on a fixed line $X$ also lies on a fixed line.

Bruh I am trying to pick up the pieces of my geometry skills First note that $\measuredangle APX=\measuredangle ADP=\measuredangle ADQ=\measuredangle AQX$, so $APXQ$ cyclic. Now we can see that $\triangle ADQ\stackrel{+}{\sim}\triangle APX$ due to the equal angles $\measuredangle APX=\measuredangle ADQ$ and $\measuredangle AQD=\measuredangle AQP=\measuredangle AXP$ (similarity thus follows from AA similarity). From here we’re almost at the end. Notice that if $X’=(APQ)\cap\overline{AD}\neq A$ then $X’$ and $X$ are reflections of each other around the perpendicular bisector of $PQ$. Further, $X’$ moves along a straight line (in fact it is the image of $\omega$ under inversion with power $\sqrt{\operatorname{Pow}_{(APQX)}(D)}$), so we are done.