A. ASA
B.SSS
C.SAS
D. ASA

Option C:
BD ≅ BD (reflexive property) and ΔABD ≅ ΔCBD (SAS).
Solution:
Given data: BD is the perpendicular bisector of AC.
In ΔABD and ΔCBD,
AB ≅ BC (BD is the bisector of AC)
∠ABD ≅ ∠CBD (Each 90°)
BD ≅ BD (reflexive property)
Therefore, ΔABD ≅ ΔCBD by SAS similarity theorem.
Hence option C is the correct answer.
BD ≅ BD (reflexive property) and ΔABD ≅ ΔCBD (SAS).
ΔABD and ΔCBD congruence by SAS theorem.
Hence proved.