congruence - SSS theoremIll send a picture of the question

From the figure,
We need to prove that,
[tex]\Delta\text{MNO}\cong\Delta\text{PQR}[/tex]By SSS theorem,
[tex]\begin{gathered} MN\cong PQ(\text{Given)} \\ NO\cong QR(\text{Given)} \end{gathered}[/tex]So, the third side must be,
[tex]MO\cong PR[/tex]Hence, the correct option is D.