CX is angle bisector of <ACB if C,X,Y are collinear, touchpoints of incircle

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CX is angle bisector of <ACB if C,X,Y are collinear, touchpoints of incircle

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#1 h Jun 9, 2020, 5:47 AM

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The incircle of the triangle $ABC$ touches the sides $AC$ and $BC$ at points $P$ and $Q$ respectively. $N$ and $M$ are the midpoints of $AC$ and $BC$ respectively. Let $X=AM\cap BP, Y=BN\cap AQ$ . Given $C,X,Y$ are collinear, prove that $CX$ is the angle bisector of the angle $ACB$ .

I. Gorodnin

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Jjesus

#2 h Jun 16, 2020, 2:17 PM • 3 Y

Y by Mango247, Mango247, Mango247

Barycentric coordinates
Be I: incenter de $\triangle ABC$ and $p=\frac{a+b+c}{2}$
$\Rightarrow M=(0,1,1)$ , $N=(1,0,1)$ , $P=(p-c,0,p-a)$ , $Q=(0,p-c,p-b)$
$\Rightarrow X=(p-c,p-a,p-a)$ , $Y=(p-b,p-c,p-b)$
Given the $C,X,Y$ are collinear $\Rightarrow a^2+b^2=c(a+b)$
$C,X,I$ are collinear if and only if $(p-c)b=(p-a)a \leftrightarrow (a+b-c)b=(b+c-a)a \leftrightarrow a^2+b^2=c(a+b)$ QED

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