We know that the sum of supplementary angels is 180 degrees.
[tex]m\angle1+m\angle2=180^o[/tex]Using triangle sum property, we get
[tex]\begin{gathered} m\angle2+\angle m3+m\angle4=180^o_{} \\ ^{} \end{gathered}[/tex][tex]\text{ Substitute }180^o=m\angle1+m\angle2,\text{ we get}[/tex][tex]\begin{gathered} m\angle2+\angle m3+m\angle4=m\angle1+m\angle2 \\ ^{} \end{gathered}[/tex][tex]\text{ Subtracting m}\angle2\text{ from both sides, we get}[/tex][tex]\begin{gathered} m\angle2+\angle m3+m\angle4-m\angle2=m\angle1+m\angle2-m\angle2 \\ ^{} \end{gathered}[/tex][tex]\begin{gathered} \angle m3+m\angle4=m\angle1 \\ ^{} \end{gathered}[/tex]We get results by using the following equations. Mia's claim must be true from the following equations.
[tex]m\angle1+m\angle2=180^o[/tex][tex]\begin{gathered} m\angle2+\angle m3+m\angle4=180^o_{} \\ ^{} \end{gathered}[/tex]Hence the answer is D.