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There exists $\triangle{ABC}$ with $AB=c, BC=a, CA=b$. Let I be the incenter and G be the centroid of $\triangle{ABC}$. Assume that $GI$ perpendicular to $CI$. Prove that $6ab=(a+b)(a+b+c)$.

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Let $D$ and $E$ be the intersections of $GI$ with $CA$ and $CB$, respectively.

If $CI\perp GI$, we must have $CD=CE$. Since the trilinear coordinates of $I$ and $G$ are $\left(1,1,1\right)$ and $\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)$, the trilinear coordinates of $D$ and $E$ are

$$ D=(cb-ca,0,ab-ac),\qquad E=(0,ac-bc,ba-bc), $$

and $CD=CE$ is equivalent to $d(D,BC)=d(E,AC)$. Given the previous line,

$$ d(D,BC) = c(b-a)\cdot\frac{2\Delta}{ac(b-a)+ac(b-c)},$$

$$ d(E,AC) = c(a-b)\cdot\frac{2\Delta}{bc(a-b)+cb(a-c)},$$

hence $d(D,BC)=d(E,AC)$ is equivalent to:

$$ ac(b-a)+ac(b-c) = bc(b-a)+bc(c-a) $$

or to:

$$ 4ab = a^2+ac+b^2+bc $$

or to:

$$ \color{red}{6ab} = (a+b)^2 + c(a+b) = \color{red}{(a+b)(a+b+c)}$$

as wanted.

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