shohvanilu wrote: In triangle $ABC$ $CL$ is a bisector($L$ lies $AB$) $I$ is center incircle of $ABC$. $G$ is intersection medians. If $a=BC, b=AC, c=AB$ and $CL\perp GI$ then prove that $\frac{a+b+c}{3}=\frac{2ab}{a+b}$ We know that $ \begin{cases} \vec{CG} = \frac{1}{3}(\vec{CA}+\vec{CB}) \\ a.\vec{IA}+b.\vec{IB}+c.\vec{IC}=\vec{0} \end{cases} $ Then by short computation we find that $ \vec{GI} = \vec{CI}-\vec{CG} = (\frac{a}{a+b+c}-\frac{1}{3})\vec{CA}+(\frac{b}{a+b+c}-\frac{1}{3})\vec{CB} $ Now $ GI \perp CI \iff \vec{GI} .\vec{CI} =0 \iff [(2a-b-c)\vec{CA} +(2b-a-c)\vec{CB} ](a.\vec{CA}+b.\vec{CB}) =0$ $ \iff (ab+\vec{CA}.\vec{CB})[b(2a-b-c)+a(2b-a-c)] = 0$ Since $ab+\vec{CA}.\vec{CB}=ab(1+ \cos C ) > 0 $ $ \iff b(3a-a-b-c)+a(3b-a-b-c) = 0 $ $ \iff 6ab=(a+b)(a+b+c) $ $ \iff \boxed{\frac{a+b+c}{3}=\frac{2ab}{a+b}}$

Let $IG \cap AC=X$ $IG \cap BC=Y$ Clearly CY=CX. Applying sine rule to $\triangle up BGY$ we get $BY/sin(\angle BGY)=BG/cos (C/2)$ Similarly applying sine rule to $\triangle up XGL$ we get $XL/sin(\angle XGL)=GL/cos (C/2)$ Dividing these relations we get $BY/XL=2$ $\Rightarrow (a-CY)/(CX-b/2)=2$ $\Rightarrow CY=CX=(a+b)/3$ Let $[XYZ]$ denote the area of $\triangle up XYZ$. $[ACL]+[BCL]=[ABC]$ $\Rightarrow CLsin(C/2)(a+b)=absinC$ $\Rightarrow CL=2abcos(C/2)/(a+b)$ Now since $AI$ is the angle bisector of $\triangle up ALC$ we have $LI/IC=AC/AL=b/(bc/a+b)=(a+b)/c$ $\Rightarrow AI=(a+b)/(a+b+c)*CL=2abcos(C/2)/(a+b+c)$ $\Rightarrow CX=CY=2ab/(a+b+c)$ Equating these relations for CX and CY we get $(a+b+c)/3=2ab/(a+b)$