In a triangle prove that $IG \perp BC \iff b + c = 3 a$ Consider the triangle $ABC$. Denote the incenter with $I$, the centroid with $G$ and $AB = c$, $AC = b$, $BC = a$.
Prove that $IG \perp BC \iff b + c = 3 a$.
I have tried using centroid's and incenter's properties but to no result. Any help is appreciated.
 A: This is a very well known problem which can be solved using vectors or Ptolemy's theorem. By the way, the question has to be worded more clearly. Please note if $b = c$, $IG \perp BC$ and then $b+c = 3a$ is not necessarily true. So the question should ask to show $b + c = 3a$, if $b \ne c$ and $IG \perp BC$.

In $\triangle ABC$, $I$ is the incenter, $G$ is the centroid and $D, E, F$ are midpoints of sides $AB, BC, CA$ resp. We draw circumcircle of $\triangle ABC$. As $AM$ is angle bisector of $\angle A$, arc $BM = MC$. As $E$ is the midpoint of chord $BC$, $ME \perp BC$ (as arc $BM = MC$).
That leads to $ME \parallel IG$ and $\triangle AIG \sim \triangle AME \implies AI = 2 MI$.
In $\triangle BMI, \angle MBI = \angle MBC + \angle CBI = \frac{\angle A + \angle B}{2} = \angle MIB \implies BM = MI$ and we already have $BM = MC$.
Given $ABCM$ is cyclic, we use Ptolemy's theorem,
$AM \cdot BC = AC \cdot BM + AB \cdot MC$
$3 MI \cdot a = MI \cdot (b + c) \implies 3a = b + c$
A: Cooment:Considering comments by cosmo5 and dnfu I consider especial case in which the triangle is isosceles.

As can be seen in figure when $IG\perp AB$ it means the triangle is isosceles; and AC=CB, then we have:
$b+c=BF+AF+AD+DC$
$DC=CE$, $BE=BF=AC=AD\rightarrow b+c=3BE+EC$
Now we have to prove that $EC=3BE$ or $(a=BC)=4BE$ such that $b+c=3(BE+EC=a)$.
,
A: Let's first observe that since $IG \perp BC$, the following relation holds between the sides of the concave quadrilateral $CIBG$:
$$IC^2+GB^2=IB^2+GC^2 \;\;\;\;\;(1)$$
If we find all of these distances in terms of the sides of the triangle (i.e. $a,b ,$ and $c$) we will be done.
To calculate the lengths of $GB^2$ and $GC^2$ we can use Apollonius's theorem and the fact that the centroid divides medians in the ratio $2:1$,
$$  GB^2=\frac{2a^2+2c^2-b^2}{9}.$$
Similarly:
$$  GC^2=\frac{2b^2+2a^2-c^2}{9}$$
To calculate the lengths of $IC^2$ and $IB^2$ we can apply the Pythagorean Theorem, but the expression for the inradius won't play out nicely, unless we get help from of our computer. Alternatively, if the following equality is known, $$\cos \left(\frac{A}{2}\right)=\sqrt{\frac{s\left(s-a\right)}{2}}$$
it is easier to see that the following results hold:
$$  IC^2=\frac{a c (s - b)}{s}$$
Similarly:
$$  IB^2=\frac{a b (s - c)}{s}$$
Substituting these expressions into $Eq. (1)$:
$$\frac{c^2-b^2}{3}=a(c-b)$$
$$(c+b)(c-b)=3a(c-b)$$
which gives us both the isosceles case and the relation given in the problem statement.

A: 
Drop a perpendicular from point $I$ on side $BC$ and let the feet of this perpendicular be $D$. $G$, $I$ and $D$ will be collinear. Drop a perpendicular from vertex $A$ on side $BC$ and let the feet of this perpendicular be $L$.
Observe that, in $\triangle ALA'$, $GD\parallel AL$ and thereafter $A'L=3A'D$.
$A'L=CL-CA'=bcos\left(\angle C\right)-\frac {a}{2}=\frac {a^{2}+b^{2}-c^{2}}{2a}-\frac {a}{2}=\frac {b^{2}-c^{2}}{2a}$
$A'D=A'B-BD=\frac {a}{2}-\left(s-b\right)$
Now, $3\{\frac {a}{2}-\left(s-b\right)\}=\frac {b^{2}-c^{2}}{2a}$ and from here simplify to get [the simplification is fairly short] $3a\left(b-c\right)=b^{2}-c^{2}$.
When $\triangle ABC$ is isosceles, then the case evidently holds but for non-isosceles triangles the $(b-c)$ cancels out and hence $b+c=3a$.
