AE//PO if PC=2AP, PB=BC, incircle (2019 All-Siberian Qualifying 9.4)

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AE//PO if PC=2AP, PB=BC, incircle (2019 All-Siberian Qualifying 9.4)

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#1 h May 17, 2021, 3:01 PM

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On the side $AC$ of triangle $ABC$ , point $P$ is chosen such that $PC = 2AP$ . Point $O$ is the center of the inscribed circle of the triangle $PBC$ , $E$ is the touchpoint of this circle with the straight line $PB$ . It turned out that $PB=BC$ . Prove that line $AE$ is parallel to line $PO$ .

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#2 h May 18, 2021, 7:59 PM

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Let $A(0,0),B(2c,b),P(c,0),C(3c,0)$ .

Incenter of $\triangle BCP\ :\ O(2c,\frac{c(\sqrt{b^{2}+c^{2}}-c)}{b})$ .

Slope of the line $PO\ :\ m_{PO}=\frac{\sqrt{b^{2}+c^{2}}-c}{b}$ .
The line $BP$ is a tangent line, point $E(\frac{c(\sqrt{b^{2}+c^{2}}+c)}{\sqrt{b^{2}+c^{2}}},\frac{bc}{\sqrt{b^{2}+c^{2}}})$ .
Slope of the line $AE\ :\ m_{AE}=\frac{\sqrt{b^{2}+c^{2}}-c}{b}$ .

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