Answer:
Δ BCD is an isosceles triangle
Explanation:
the exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.
∠ ABC is an exterior angle of the triangle , then
∠ BDC + ∠ BCD = ∠ ABC
62° + ∠ BCD = 124° ( subtract 62° from both sides )
∠ BCD = 62°
then ∠ BDC = ∠ BCD = 62°
thus the 2 base angles of the triangle are congruent
then triangle BCD is isosceles with BC = BD