Given that
[tex]\bar{AD}\cong\bar{AE}[/tex]
Therefore,
[tex]\begin{gathered} m\angle BAD\cong m\angle ADE=62^0 \\ m\angle BAD\cong m\angle AED=62^0 \\ m\angle DAE=? \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} m\angle ADE+m\angle AED+m\angle DAE=180^0 \\ 62^0+62^0+m\angle DAE=180^0 \\ 124^0+m\angle DAE=180^0 \\ m\angle DAE=180^0-124^0 \\ m\angle DAE=56^0 \end{gathered}[/tex]
Therefore,
[tex]m\angle DAE=56^0[/tex]
Reason
Sum of angles in a triangle.
