Answer:
Δ ABC is isosceles
Explanation:
∠ ABC and ∠ BCE are alternate angles and are congruent , that is
∠ BCE = ∠ ABC = 80°
since DE id a straight line then the angles lying on it sum to 180° , that is
∠ ACD + ∠ ACB + ∠ BCE = 180°
50° + ∠ ACB + 80° = 180°
∠ ACB + 130° = 180° ( subtract 130° from both sides )
∠ ACB = 50°
∠ BAC and ∠ ACD are alternate angles and are congruent , that is
∠ BAC = 50°
then
∠ BAC = ∠ ACB = 50°
Since the 2 base angles are congruent then Δ ABC is isosceles