Problem Let $AD, \ BE$, and $CF$ be the three medians of $△ABC$, and $G$ be their intersection point. Then, $$AD+BE+CF<AB+BC+CA<2 \ (AG+BG+CG).$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution First, from the problem $0098$, $$AD+BE+CF<AB+BC+CA. \qquad [1]$$…

Problem Let $AD, \ BE$ and $CF$ be the three median lines of $△ABC$, respectively. Then, $$\frac{1}{2} (AB+BC+CA)<AD+BE+CF<AB+BC+CA.$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution From the problem $0097 \ (2)$, $$\frac{1}{2} (AB+CA-BC)<AD, \qquad \frac{1}{2} (AB+BC-CA)<BE,$$ $$and \qquad…

Problem If the midpoint of the side $BC$ of $△ABC$ is $D$, then $$(1) \ AD<\frac{1}{2} (AB+AC).$$ $$(2) \ AD>\frac{1}{2} (AB+AC-BC).$$ $$ $$ $$ $$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $\downarrow$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ $$ Solution $(1)$ If we extend $AD$ and take a point $E$…