Cho tam giác ABC với đường cao \({h_a},{h_b},{h_c}\) thỏa mãn \(\frac{{{h_a}}}{{{h_b}}} + \frac{{{h_b}}}{{{h_c}}} + \frac{{{h_c}}}{{{h_a}}} = \frac{{{h_b}}}{{{h_a}}} + \frac{{{h_c}}}{{{h_b}}} + \frac{{{h_a}}}{{{h_c}}}\). Chứng minh rằng tam giác ABC là tam giác cân.

Ta có: \(S = \frac{1}{2}a.{h_a} = \frac{1}{2}b.{h_b} = \frac{1}{2}c.{h_c}\) nên \({h_a} = \frac{{2S}}{a};{h_b} = \frac{{2S}}{b},{h_c} = \frac{{2S}}{c}\)

Do đó: \(\frac{{{h_a}}}{{{h_b}}} + \frac{{{h_b}}}{{{h_c}}} + \frac{{{h_c}}}{{{h_a}}} = \frac{{{h_b}}}{{{h_a}}} + \frac{{{h_c}}}{{{h_b}}} + \frac{{{h_a}}}{{{h_c}}}\) suy ra \(\frac{b}{a} + \frac{c}{b} + \frac{a}{c} = \frac{a}{b} + \frac{b}{c} + \frac{c}{a}\)

\({b^2}c + a{c^2} + {a^2}b = {a^2}c + b{c^2} + a{b^2}bc\left( {b - c} \right) - a\left( {{b^2} - {c^2}} \right) + {a^2}\left( {b - c} \right) = 0\)

\(\left( {b - c} \right)\left( {bc - ab - ac + {a^2}} \right) = 0\)

\(\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right) = 0\)

Suy ra \(\left[ \begin{array}{l}a = b\\b = c\\c = a\end{array} \right.\). Suy ra, tam giác ABC cân.