Cho tam giác $ABC$ có : $\frac{r}{r_a} = \frac{1}{3};\frac{r}{R} = \frac{2}{5}$. Chứng minh rằng tam giác $ABC$ vuông.
$\frac{r}{{{r_a}}} = \frac{1}{3} \Rightarrow \frac{{p - a}}{p} = \frac{1}{3} \Rightarrow 3\left( {p - a} \right) = p \Rightarrow 2p = 3a$
    $\begin{array}{l}
 \Rightarrow a + b + c = 3a \Rightarrow b + c = 2a &  &  & \left( 1 \right)\\
\frac{r}{R} = \frac{2}{5} \Rightarrow \frac{{\frac{S}{p}}}{{\frac{{abc}}{{4S}}}} = \frac{2}{5} \Rightarrow \frac{{4{S^2}}}{{pabc}} = \frac{2}{5}\\
 \Rightarrow \frac{{4p\left( {p - a} \right)\left( {p - b} \right)\left( {p - c} \right)}}{{pabc}} = \frac{2}{5}\\
 \Rightarrow \frac{{\frac{1}{2}\left( {b + c - a} \right)\left( {c + a - b} \right)\left( {a + b - c} \right)}}{{abc}} = \frac{2}{5}
\end{array}$
Kết hợp với (1) ta có:
   
$\begin{array}{l}
\frac{{\frac{1}{2}a\frac{{3c - b}}{2}.\frac{{3b - c}}{2}}}{{abc}} = \frac{2}{5} \Rightarrow \frac{{\left( {3c - b} \right)\left( {3b - c} \right)}}{{8bc}} = \frac{2}{5}\\
 \Rightarrow \frac{{10bc - 3\left( {{b^2} + {c^2}} \right)}}{{bc}} = \frac{{16}}{5} \Rightarrow \frac{{16bc - 3{{\left( {b + c} \right)}^2}}}{{bc}} = \frac{{16}}{5}\\
 \Rightarrow 16 - 12.\frac{{{a^2}}}{{bc}} = \frac{{16}}{5} \Rightarrow \frac{{3{a^2}}}{{bc}} = \frac{{16}}{5}\\
 \Rightarrow \frac{b}{a}.\frac{c}{a} = \frac{{15}}{{16}}      \left( 2 \right)
\end{array}$
Từ (1) $ \Rightarrow \frac{b}{a} + \frac{c}{a} = 2    \left( 3 \right)$
Từ (2), (3) suy ra: $\frac{b}{a},\frac{c}{a}$ là hai nghiệm của phương trình bậc hai
    ${X^2} - 2X + \frac{{15}}{{16}} = 0.$ Vậy: $\left[ \begin{array}{l}
\left\{ \begin{array}{l}
\frac{b}{a} = \frac{5}{4}\\
\frac{c}{a} = \frac{3}{4}
\end{array} \right.\\
\left\{ \begin{array}{l}
\frac{b}{a} = \frac{3}{4}\\
\frac{c}{a} = \frac{5}{4}
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\frac{a}{4} = \frac{b}{5} = \frac{c}{3}\\
\frac{a}{4} = \frac{b}{3} = \frac{c}{5}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
B = {90^\circ}\\
C = {90^\circ }
\end{array} \right.  (\text{Theo định lí Pitago} ) $


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