Final answer:
To prove that ∠ABD is an acute angle, we can use the properties of isosceles triangles and exterior angles. Given that CD = CB, we can conclude that triangle BCD is isosceles, and ∠DBC is an acute angle. Since ∠ABD is an exterior angle of triangle BCD, its measure is the sum of the measures of ∠DBC and ∠DCB, which are both acute angles. Therefore, ∠ABD is an acute angle.
Step-by-step explanation:
To prove that ∠ABD is an acute angle, we need to show that its measure is less than 90 degrees. Given that CD = CB, we can conclude that triangle BCD is an isosceles triangle, and therefore ∠DBC ≅ ∠DCB. Since BC > AC and ∠DCB is an acute angle, it follows that ∠DBC is also an acute angle. Now, ∠ABD is an exterior angle of triangle BCD, so its measure is equal to the sum of the measures of ∠DBC and ∠DCB. Since both of these angles are acute, their sum is also an acute angle. Therefore, ∠ABD is an acute angle.