Respuesta :

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.

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