Inside an isosceles triangle $ABC$, right at $B$, the point $P$ is marked in such a way that $\dfrac{PC}{1}=\dfrac{PB}{2}=\dfrac{PA}{3}.$ Calculate $\angle BPC.$
I found a solution but I have some difficulties in understanding that. Several considerations were made that I found no basis.
Draw $\triangle BCE \cong \triangle BCP$ and $\triangle ACF \cong \triangle APC$
Connect $EP \implies \triangle PCE$ (isosceles) and $\angle PEC=CPE = 67.5^\circ$
$\angle BPD = \angle DPA$
$DB \parallel EP \implies \angle BPE = \angle BPD$
In $\triangle BPA$, $45^\circ-VE+90^\circ-VE+4VE = 180^\circ \\ \therefore VE = 22.5^\circ\implies \angle BPE = 67.5^\circ$
Therefore $\angle BPC = 2\angle BPE = 135^\circ$.
Add another congruent triangle $BCD$ and put point $E$ in it such that $DE=CP, ~ BE=BP, ~ CE=AE.$ Then $\angle PBE = 90°,$ $PE = 2x\sqrt{2}$. Then $\angle CPE = 90°.$ Therefore $\angle BPC = 135°.$