答案
(1)解:∵AB∥CD,∠A=37°,
∴∠ECD=∠A=37°,
∵DE⊥AE,
∴∠D=90°-∠ECD=90°-37°=53°.
(2)证明:∵AB=AC,AD=AE,∴∠B=∠C,∠ADE=∠AED,
∴∠ADE+∠DAE=∠AED+∠DAE,即∠ADB=∠AEC,
∴△ABD≌△ACE,
∴BD=CE
(1)解:∵AB∥CD,∠A=37°,
∴∠ECD=∠A=37°,
∵DE⊥AE,
∴∠D=90°-∠ECD=90°-37°=53°.
(2)证明:∵AB=AC,AD=AE,∴∠B=∠C,∠ADE=∠AED,
∴∠ADE+∠DAE=∠AED+∠DAE,即∠ADB=∠AEC,
∴△ABD≌△ACE,
∴BD=CE