GARA di FEBBRAIO

Let $P'$ be the reflection of point $P$ with respect to line $BC$. Note that points $A$, $H$, $P'$ are collinear, since $\widehat{AHP'}=\widehat{AHP}+2\widehat{PHC}=\widehat{AHP}+2\left(90^\circ-\widehat{AHP}/2\right)=180^\circ$.

Now, $\widehat{AP'C}=\widehat{HP'C}=\widehat{HPC}$ by symmetry, but $\widehat{HPC}=\widehat{ABC}$ by hypothesis; this implies the cyclicity of quadrilateral $ABP'C$.

This cyclicity implies that $\widehat{BAC}$ is supplementary to $\widehat{BP'C}$, which is congruent to $\widehat{BPC}$ by symmetry, and hence $\widehat{BAC}=180^\circ-\widehat{BPC}=180^\circ-130^\circ=50^\circ$.

Second solution. Let $C'$ be the intersection of line $CP$ with side $AB$; quadrilateral $BHP C'$ is cyclic (since $\widehat{HPC}=\widehat{HBC'}$ by hypothesis, and thus the opposite angles $\widehat{HBC'}$ and $\widehat{C'PH}$ are supplementary).

Note moreover that, denoting by $2\theta$ the angle $\widehat{PHA}$, we have $\widehat{PHC}=90^\circ-\theta=\widehat{C'HB}$; using again the identity $\widehat{HPC}=\widehat{HBC'}$ in triangles $BAH$ and $PCH$, it follows by difference that $\widehat{BAH}=\widehat{PCH}=\widehat{C'CH}$. It follows that quadrilateral $AC'H C$ is also cyclic.

There are now various ways to conclude using angle identities given by cyclicity; for example, $\widehat{HBP}=\widehat{HC'P}$ by cyclicity of $BHPC'$, and $\widehat{HC'P}=\widehat{HAC}$ by cyclicity of $AC'HC$; but then $\widehat{BAC}=\widehat{BAH}+\widehat{HAC}=\widehat{PCB}+\widehat{PBC}=180^\circ-\widehat{BPC}=50^\circ$.