Question

Answer 1

  

Answer 2

Ta có:
\(S_{ABC}=\dfrac{1}{2}h_a.a\)

\(S_{ABC}=\dfrac{1}{2}b.c.sinA\)

\(\Rightarrow\left\{{}\begin{matrix}h_a=\dfrac{2S_{ABC}}{a}\\c.sinA=\dfrac{2S_{ABC}}{b}\end{matrix}\right.\)

Để \(h_a=c.sinA\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\) Tam giác ABC cân tại C