# Find $\angle BPC$ in the triangle below

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 $$\angle BPC = 2\angle BPE = 135^\circ$$.

• Why do you think CP is angle bisector? Commented Jan 26, 2022 at 13:48
• Angles BPE and EPC cannot be equal. If they are equal then triangles BPE and CPE are similar. They have common side PE, then triangles BPE and CPE are congruent. Then $2x=x$. Contradiction. Commented Jan 26, 2022 at 13:58
• Also the last line would be $135°$, switched a couple digits Commented Jan 26, 2022 at 14:04
• @IvanKaznacheyeu I think you didn't understand...it wasn't my resolution but the resolution I found and I didn't understand some considerations made...so I posted here to find another solution or an explanation of the resolution Commented Jan 26, 2022 at 14:33

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 $$\angle BPC = 135°.$$
• Why $\angle PBE=90$? Commented Jan 26, 2022 at 14:38