Đáp án:
$S=\dfrac{1}{2}bc\quad sin A=\dfrac{1}{2}ah_a$
a.sinA + b.sinB + c.sinC = ha + hb + hc
$→$ $a.\dfrac{2S}{bc}+b.\dfrac{2S}{ac}+c.\dfrac{2S}{ab}=\dfrac{2S}{a}+\dfrac{2S}{b}+\dfrac{2S}{c}$
$→(a-b)^2+(b-c)^2+(a-c)^2=0$
mà $(a-b)^2+(b-c)^2+(a-c)^2≥0$
$→a=b=c$
$→Δ$ đã cho là $Δ$ đều