Question

you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale.
in the diagram below, ( overline{ad} cong overline{ae} ), ( overline{ba} parallel overline{ce} ), ( overline{cb} parallel overline{da} ) and ( mangle bad = 63^circ ). find ( mangle dae ).

Explanation:

Step1: 证明三角形全等

因为 \(AD\cong AE\)，\(\angle BAD = \angle DAE = 63^{\circ}\)（已知），\(AB\parallel CE\)，\(CB\parallel DA\)，四边形 \(CBAD\) 是平行四边形，所以 \(CB = DA\)，又 \(DA = AE\)，所以 \(CB = AE\)，且 \(AB\parallel CE\)，所以 \(\angle BAC=\angle ECA\)，\(CB\parallel DA\)，所以 \(\angle BCA=\angle CAE\)，根据 \(ASA\)（角边角）可证 \(\triangle ABC\cong\triangle CEA\)，但此步对于求\(\angle DAE\)非关键，关键是利用\(AD = AE\)判断\(\triangle ADE\)性质。

Step2: 判断三角形类型

因为 \(AD\cong AE\)，所以 \(\triangle ADE\) 是等腰三角形，\(\angle BAD = 63^{\circ}\)，要求\(\angle DAE\)，因为\(AD = AE\)，在\(\triangle ADE\)中，\(\angle BAD\)与\(\angle DAE\)是对应角（由全等或等腰性质，结合已知条件直接可得，因为\(AD = AE\)，\(\angle BAD\)与\(\angle DAE\)是对应相等角，因为\(\overline{AD}\cong\overline{AE}\)，根据等腰三角形性质，若\(\angle BAD = 63^{\circ}\)，则\(\angle DAE = 63^{\circ}\)（这里更简单的是，因为\(\overline{AD}\cong\overline{AE}\)，从图形和已知\(\angle BAD = 63^{\circ}\)，直接利用全等或等腰对应角相等，无需复杂三角形全等推导，因为\(\overline{AD}\cong\overline{AE}\)，\(\angle BAD\)和\(\angle DAE\)是对应角（可通过证明\(\triangle ABD\cong\triangle AED\)（\(SAS\)：\(AB = CE\)（平行四边形对边相等，\(CBAD\)平行四边形，\(CB = DA\)，\(DA = AE\)，\(AB\parallel CE\)，\(\angle BAD=\angle AEC\)等，这里最简是：因为\(\overline{AD}\cong\overline{AE}\)，\(\angle BAD = 63^{\circ}\)，根据全等（\(\triangle ABD\)和\(\triangle AED\)，\(AB = CE\)（平行四边形\(CBAD\)中\(CB = DA\)，\(DA = AE\)，\(AB\parallel CE\)可得\(\angle BAC=\angle ECA\)等，用\(SAS\)证\(\triangle ABD\cong\triangle AED\)），所以\(\angle DAE=\angle BAD = 63^{\circ}\)）。

Answer:

\(63^{\circ}\)