We have AB = AC …… (i)
AD = DA (common) …… (ii)
And, ∠ADC = ∠ADB (AD ⊥ BC at point D) …… (iii)
Therefore, from (i), (ii) and (iii), by RHS congruence condition, ΔABD ≅ ΔACD, the triangles are congruent.
Therefore, BD = CD.
And ∠ABD = ∠ACD (corresponding parts of congruent triangles)