Provide the requested proof.
-$$\text { In the trapezoid below, } \mathrm { m } ( \angle \mathrm { D } ) = \mathrm { m } ( \angle \mathrm { C } ) \text {. Prove that } \mathrm { AD } = \mathrm { BC } \text {. }$$

Question

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The Answer is Answers will vary. A sample correct response would be as follows.
Mark points E and F on DC such that AE and BF are perpendicular to DC. $Answers will vary. A sample correct response would be as follows. Mark points E and F on DC such that AE and BF are perpendicular to DC. As \mathrm { ABFE } is a parallelogram (in fact, a rectangle), \mathrm { AE } = \mathrm { BF } . By \mathrm { AAS } , riangle \mathrm { ADE } \simeq riangle \mathrm { BCF } , an hence A D = B C .$
As

\mathrm { ABFE }

is a parallelogram (in fact, a rectangle),

\mathrm { AE } = \mathrm { BF }

. By

\mathrm { AAS } , riangle \mathrm { ADE } \simeq riangle \mathrm { BCF }

, an hence

A D = B C

.

Provide the Requested Proof