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Let $ABC$ be an acute angled triangle whose incenter and centroid are respectively $I$ and $G$.$AI,BI$ and $CI$ cuts the sides of the triangle at $P,Q,R$ respectively.If $p_1,p_2$ and $p_3$ are the lengths of the altitudes through $A,B$ and $C$ respectively and $G$ lies on $PQ$ then prove by vector method that $p_1+p_2=p_3$.

I could not much do about this complex problem which i could mention here.I know that incenter,orthocenter and centroid all lie on one line

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    $\begingroup$ incenter, orthocenter and centroid all lie on one line - No, they don't. You're confusing incenter with circumcenter. $\endgroup$ – Lucian Aug 19 '15 at 11:14
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Using trilinear coordinates, we have $I=[1;1;1]$ and $G=\left[\frac{1}{a},\frac{1}{b},\frac{1}{c}\right]=[h_a;h_b;h_c]$, so: $$ P=[0;1,1],\quad Q=[1;0;1] $$ and $G$ lies on $PQ$ iff $\det[G,P,Q]=0$. Since: $$\det\begin{pmatrix}h_a & h_b & h_c \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{pmatrix}=h_a+h_b-h_c$$ the claim trivially follows.

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  • $\begingroup$ How are the trilinear coordinates of P, Q and G found? $\endgroup$ – diya Aug 19 '15 at 16:20
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    $\begingroup$ @diya: Very well-known facts: the distances of $G$ from the side lengths are obviously $h_a/3,h_b/3,h_c/3$. Moreover, $A=[1,0,0]$, $I=[1,1,1]$ and $P=[0,?,?]$ have to be collinear, hence $P=[0,1,1]$. $\endgroup$ – Jack D'Aurizio Aug 19 '15 at 16:27

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