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Problema Proposto
24 - No interior de um triângulo retângulo
isósceles ABC, reto en B, se marca o ponto
"P''. de tal manera que :
[tex3]\frac{PC}{1}=\frac{PB}{2}=\frac{PA}{3}[/tex3]
Calcular [tex3]\angle[/tex3]BPC

Traçando o triângulo BCD congruente ao triângulo ABC de modo que

[tex3]\mathsf{DE=CP\\ BE=BP\\ CE=AE\\
\therefore ∠PBE=90°\implies PE=2x\sqrt2\\
\therefore ∠CPE=90°(\triangle DBE \cong \triangle CBP \implies \angle PBE = \angle PBC + \angle CBE = \angle EBD + \angle CBE = 90°.\\
\therefore \boxed{\color{red}∠BPC=135°}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{DE=CP\\ BE=BP\\ CE=AE\\
\therefore ∠PBE=90°\implies PE=2x\sqrt2\\
\therefore ∠CPE=90°(\triangle DBE \cong \triangle CBP \implies \angle PBE = \angle PBC + \angle CBE = \angle EBD + \angle CBE = 90°.\\
\therefore \boxed{\color{red}∠BPC=135°}}[/tex3]

(Solução: Ivan K.)