Provide the requested proof.
-In the trapezoid below, $\mathrm { m } ( \angle \mathrm { D } ) = \mathrm { m } ( \angle \mathrm { C } )$ and $\mathrm { AD } = \mathrm { BC }$. If $\mathrm { M }$ is the midpoint of $\overline { \mathrm { CD } }$, prove that $\mathrm { m } ( \angle \mathrm { MAB } ) = \mathrm { m } ( \angle \mathrm { MBA } )$.

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The Answer of Provide the requested proof.
-In the trapezoid below, $\mathrm { m } ( \angle \mathrm { D } ) = \mathrm { m } ( \angle \mathrm { C } )$ and $\mathrm { AD } = \mathrm { BC }$. If $\mathrm { M }$ is the midpoint of $\overline { \mathrm { CD } }$, prove that $\mathrm { m } ( \angle \mathrm { MAB } ) = \mathrm { m } ( \angle \mathrm { MBA } )$.

Provide the Requested Proof m(∠D)=m(∠C)\mathrm { m } ( \angle \mathrm { D } ) = \mathrm { m } ( \angle \mathrm { C } )m(∠D)=m(∠C)

Provide the Requested Proof $\mathrm { m } ( \angle \mathrm { D } ) = \mathrm { m } ( \angle \mathrm { C } )$