Cho tam giác ABC có cạnh a, b, c thỏa mãn bc = a².

Chứng minh rằng sinB.sinC = sin²A và h_b . h_c = h_a².

• Ta có: \(\frac{{\sin {\rm{A}}}}{a} = \frac{{\sin B}}{b} = \frac{{\sin C}}{c} = 2{\rm{R}}\)

Suy ra \(a = \frac{{\sin B}}{{2{\rm{R}}}};\,\,b = \frac{{\sin {\rm{A}}}}{{2{\rm{R}}}};\,\,c = \frac{{\sin C}}{{2{\rm{R}}}}\)

Mà bc = a²

Suy ra \(\frac{{\sin B}}{{2{\rm{R}}}}.\frac{{\sin C}}{{2{\rm{R}}}} = {\left( {\frac{{\sin {\rm{A}}}}{{2{\rm{R}}}}} \right)^2}\)\( \Leftrightarrow \frac{{\sin B.\sin C}}{{4{{\rm{R}}^2}}} = \frac{{{{\sin }^2}A}}{{4{{\rm{R}}^2}}}\)

Do đó sin B . sin C = sin²A

• Ta có: \[S = \frac{1}{2}a{h_a} = \frac{1}{2}b{h_b} = \frac{1}{2}c{h_c}\]

Suy ra \[a = \frac{{2{\rm{S}}}}{{{h_a}}};b = \frac{{2{\rm{S}}}}{{{h_b}}};c = \frac{{2{\rm{S}}}}{{{h_c}}}\]

Mà bc = a²

Suy ra \[{\left( {\frac{{2{\rm{S}}}}{{{h_a}}}} \right)^2} = \frac{{2{\rm{S}}}}{{{h_b}}}.\frac{{2{\rm{S}}}}{{{h_c}}}\]

\[ \Leftrightarrow \frac{{4{{\rm{S}}^2}}}{{{h_a}^2}} = \frac{{{\rm{4}}{{\rm{S}}^2}}}{{{h_b}.{h_c}}}\]

Do đó h_b . h_c = h_a².