cos( 180 -a)=-cosa=(m^2+ 16/3-bc^2)/(8√3m/3)

The sum of the two formulas is 4m 2+16/3-4+2m 2+32/3-2bc 2 = 0.

3m^2=BC^2-6 ( 1)

From cosine theorem AC 2 = AB 2+BC 2-2AB * BC COS ∠ ABC

9m^2=4+BC^2-4BC* 1/3

27m^2= 12+3BC^2-4BC (2)

At the same time (1)(2) BC=3

2) Area of triangle ABC =( 1/2)AB*BCsin∠ABC.

=( 1/2)*2*3*2√2/3

=2√2

Area of triangle DBC =( 1/3) Area of triangle ABC =2√2/3.