Answer:
A) SSS
Step-by-step explanation:
For the given figure:
Given:
[tex]AB\cong DC[/tex]
[tex]AC\cong DB[/tex]
To prove
[tex]\triangle ABC\cong \triangle DCB[/tex]
In Δ ABC and ΔDCB
[tex]AB\cong DC[/tex] [Given]
[tex]AC\cong DB[/tex] [Given]
[tex]BC\cong CB[/tex] [ By reflexive property of congruence side BC
congruent to itself ]
[tex]\triangle ABC\cong \triangle DCB[/tex] [SSS congruence postulate]
