Assuming u mean, u can re-store $\triangle ABC$ using only ruler and compass, then yes. Clearly its known that we can draw perpendicular bisectors so we can draw the circumcenter of $\triangle ABC$, call it $O$, now we draw $AO \cap \omega=A'$, also using compass we can also draw the reflection of a point over a segment clearly, so we can draw $H_A$ which is reflection of $A'$ over $OW$, which in fact by isogonality of $AO$ and the A-altitude in $\angle BAC$ we get that $AH_A \perp BC$, now we draw $AH_A \cap PA'$, from spiral sim propeties of $P$ which is the A-queque point, we know that we just drawn the orthocenter, so now we draw the perpendicular bisector of $HH_A$ which is clearly line $BC$ and since we can draw $\omega$ we are done, thus we can restore $\triangle ABC$ and we finished

kamatadu wrote: The bisector of angle $A$ of triangle $ABC$ meet its circumcircle $\omega$ at point $W$. The circle $s$ with diameter $AH$ ($H$ is the orthocenter of $ABC$) meets $\omega$ for the second time at point $P$. Restore the triangle $ABC$ if the points $A$, $P$, $W$ are given. We have to $AP \perp PH$; Furthermore, $PH$ cuts $BC$ at its midpoint. Taking this into account, the construction of triangle $ABC$, given $A, P, W$, is immediate: $\bullet$ The center $O$ of the circle ($APW$) is constructed, which must be the circumcenter of $ABC$. $\bullet$ The perpendicular to $AP$ through $P$ cuts $OW$ at $M$ (midpoint of $BC$). $\bullet$ The points of intersection of the circle ($APW$) with the perpendicular to $OW$ through $M$ are the vertices $B$ and $C$ of the triangle to be constructed. NOTE: Once the points $A$ and $P$ are fixed, the construction of triangle $ABC$ is NOT possible if $W$ is located in the region between the perpendiculars to $AP$ in $A$ and $P$. Apple GeoGebra

first draw $\omega$ with center $O$ then draw a line through $A$ parallel to $WO,$ it intersects perp.bisector of $AP$ at midpoint of $AH$ so $H$ is constructible. next $PH$ cuts $\omega$ at $Q,$ antipode of $A.$ then take midpoint $M$ of $HQ$ also midpoint of $BC,$ then just take line through $M$ perpendicular to $WO$ to get $B,C.$