Question

E为菱形ABCD边BC上一点,AB=AE,AE交BD于点O,角DAE=2角BAE

Answer 1

证明：∵∠DAE=2∠BAE ,AD‖BC∴∠AEB=∠DAE

∵AB=AE∴∠ABE=∠AEB∴∠ABE=∠DAE=2∠BAE

∴设∠BAE=x°所以∠ABE=∠AEB=2x°

∴x+2x+2x=180,x=36°

∴∠ABE=∠AEB=∠DAE=72°

∴∠BAD=108°

∵是菱形∴AB=AD,∠ABD=∠ADB=36°∴∠DOA=72°,△DOA是等腰三角形

∴证明△ABE和△DOA全等就行

∠BAE=∠ADO,AB=DA,∠ABE=∠DAO

∴全等

∴OA=EB