Given that the side length of △ ABC is a, B, C, and (B-C) 2 + (2a + b) (C-B) = 0, try to determine the shape of △ ABC

Given that the side length of △ ABC is a, B, C, and (B-C) 2 + (2a + b) (C-B) = 0, try to determine the shape of △ ABC

The side lengths of ∵ △ ABC are a, B, C, and (B-C) 2 + (2a + b) (C-B) = 0, ∵ (B-C) [(B-C) - (2a + b)] = 0, ∵ (B-C) (b-c-2a-b) = 0, ∵ (B-C) (b-c-2a-b) = 0, ∵ B-C = 0, or - C-2A = 0, ∵ B = C or C = - 2A (rounding off), ∵ △ ABC is an isosceles triangle

If we know that the three sides of △ ABC are 5, 13 and 12, then the area of △ ABC is ()
A. 30b. 60C. 78d. Not sure

∵ 52 + 122 = 132, ∵ triangle is right triangle, ∵ length is 5, 12 sides are right sides, ∵ triangle area = 12 × 5 × 12 = 30