By straightforward complex bash, $\overline h=\frac{4i}{b-c}+\frac{\frac{b-c-2i(b+c)}{b-c}}p$ which obviously lies on a circle.

We rephrase the problem as follows: Quote: Let $B,C$ be different points on circle $\omega$. Let $P$ be a variable point on the arc $BC$ of $\omega$ not containing $A$. Let $H$ be the point such that $HX=PB, HY=PC,$ with $X,Y$ being the projections of $H$ on $PB,PC$ respectively, and $H,P$ lie on different sides of line $BC$. Then, prove that as $P$ varies, $H$ lies on a fixed circle. Let $P_1,P_2$ be the midpoints of minor and major arcs $BC$, respectively. Let $P$ coincide with $P_1,P_2$ (ignore the condition that $P$ lies on minor arc $BC$) and assume that $H_1,H_2$ are the coressponding points. Note that points $H_1,H_2$ are fixed. We claim that $H$ lies on the circle of diameter $H_1H_2$, which will evidently solve the problem. We present the following Claim. Claim: $HH_1 \parallel PP_1$. Proof: Note that $HX/HY=PB/PC,$ and so $H$ lies on the $P-$ symmedian of triangle $PBC$. Let $HP,H_1P_1$ intersect at point $X$. Moroever, let $\angle BP_2C=x$ and $M$ be the midpoint of $BC$. Then, we compute: $HX/HP=\sin \angle HPC=\sin \angle MPC,$ and so $HP=PB/\sin \angle MPC, \,\,\, (1)$. Moreover, triangles $PBX$ and $PMC$ are similar, hence $PX/PM=BX/MC=BX/BM=1/\sin \angle MXB=1/\cos x,$ hence $PX=PM/\cos x, \,\,\, (2)$. Now, in a similar manner we obtain that $H_1P_1=P_1B/\sin \angle MP_1C, \,\,\, (3)$ and $P_1X=P_1M/\cos x, \,\,\, (4)$. Now, note that $\dfrac{PX}{PH} \cdot \dfrac{P_1H_1}{P_1X}=\dfrac{PM}{P_1M} \cdot \dfrac{P_1H_1}{PH}=\dfrac{PM}{P_1M} \cdot \dfrac{P_1B}{PB} \cdot \dfrac{\sin \angle MPC}{\sin \angle MP_1C}$ However, $PM/PB \cdot \sin \angle MPC=\sin \angle PBC/\sin \angle PMC \cdot \sin \angle MPC=\sin \angle PBC \cdot MC/PC=MC/2R,$ and similarly $P_1M/P_1B \cdot \sin \angle MP_1C=MC/2R,$ and so the above expression equals $1$, hence $PX/PH=P_1X/P_1H_1,$ and so the desired result follows from Thales' theorem $\blacksquare$ Now, to the problem, in a similar manner to the above Claim we obtain that $HH_2 \parallel PP_2,$ and so $\angle H_1HH_2=\angle P_1PP_2=90^\circ$, hence $H$ belongs on the circle with diameter $H_1H_2,$ as desired.

Sol:- Construct square $BCRS$ such that $A,R,S$ lie on same side of $BC$.Let the tangent at $B$ to $(ABC)$ meet $RS$ at $L$ and the tangent at $C$ meet $RS$ at $M$.Clearly $L,M$ are fixed points. Step 1 $DE$ passes through $L$ and $FG$ passes through $M$. Proof By spiral similarity at $B$ sending square $BPDE$ to $BCRS$ we know that $\Delta BES \sim \Delta BPC$.So $\measuredangle BLS=\measuredangle LBC=\measuredangle BPC=\measuredangle BES$ so $LSEB$ is cyclic.So $\angle BEL=90^\circ=\angle BED \implies D,L,E$ are collinear. Similarly $FG$ passes through $M$ Step 2 $H$ lies on fixed circle. Proof $L,M$ are fixed points. Due to parallel lines $\measuredangle LHM=\measuredangle BPC=\measuredangle BAC$ which is a fixed angle.Since $\measuredangle LHM$ is a fixed angle subtended on fixed segment $LM$ , $(LHM)$ is a fixed circle.

Really nice problem Claim $:$ As $P$ varies all $ED$ lines are concurrent. Proof $:$ Assume $P$ and $P'$ on arc $BC$ and assume $BPDE$ and $BP'D'E'$ are squares. Let $DE$ and $D'E'$ meet at $X$. Note that $\angle BEX = \angle BE'X = \angle 90 \implies BEE'X$ is cyclic so $\angle BXE = \angle BE'E = \angle BP'P$ and since $BP || DE$ we have $\angle 180 - \angle PBX = \angle BXE = \angle BP'P \implies BX$ is tangent to $ABC$. so as $P$ varies $ED$ lines all concur at $X$. we have the same case with all $FG$ lines so assume they concur at some point $Y$. Now Note that $PDGH$ is cyclic so $\angle XHY = \angle 180 - \angle GPD$ so we have to prove $\angle GPD$ is constant which is true since $\angle GPD = \angle BPD+\angle GPC-\angle BPC = 180 - \angle BPC = \angle BAC$