[tex3]Bissetriz~BR^*: \frac{2cos(45^o)}{BR} = \frac{1}{BP}+\frac{1}{BA} \implies \\
\frac{\sqrt{2}}{BR} = \frac{1}{BP}+\frac{1}{BA}(I)\\
Por~Descartes: \frac{2}{BQ} = \frac{1}{BP}+\frac{1}{BC} = \frac{1}{BP}+\frac{1}{BA}(II)\\
I=II: \frac{2}{BQ} = \frac{\sqrt{2}}{BR}\rightarrow \begin{cases}
BQ = 2k \\
BR = k\sqrt{2}
\end{cases}\\
Lei~dos~cossenos \triangle BRQ:\\
RQ^2 = (2k)^2 + (k\sqrt{2})^2 - 2\cdot 2k \cdot k\sqrt{2} \cdot cos(45^o)\\
RQ = k\sqrt{2} \implies \triangle BRQ(isósceles) \therefore\boxed{ \color{red}x = 45^o}[/tex3]
(Solução: fornecida por Italo25 -
viewtopic.php?f=8&t=88937&p=245306&hili ... co#p245306)
*