Answer:
see explanation
Explanation:
given Δ ABD is isosceles with AB = AD then base angles are congruent, so
∠ ADB = ∠ ABD = 72°
then
∠ BAD = 180° - (72 + 72)° = 180° - 144° = 36° ( sum of angles in a triangle )
given Δ ABC is isosceles , then
∠ BCD = ∠ BAD = 36°
the exterior angle of a triangle is equal to the sum of the 2 opposite interior angles.
∠ ADB is an exterior angle of Δ BCD , then
∠ BCD + ∠ DBC = ∠ ADB
36° + ∠ DBC = 72° ( subtract 36° from both sides )
∠ DBC = 36°
So ∠ BCD = ∠ DBC = 36°
Thus the base angles of Δ BCD are congruent , then
Δ BCD is isosceles with BD = CD