Question

菱形ABCD中,E在BC上,AE交BD于M,AB等于AE,角BAE等于二分之一角EAD.求证

Answer 1

证明：∵ABCD是菱形

∴∠ABD=∠CBD

AD∥BC

∴∠AEB=∠EAD

∵2∠BAE=∠EAD

∴2∠BAE=∠AEB

∵AB=AE

∴∠ABE=∠AEB

∵∠ABE+∠AEB+∠BAE=180°

∴∠BAE=36°

∠ABE=∠AEB=72°

∴∠ABD=∠CBD=36°

∵∠BME=∠BAE+∠ABD=72°

∴∠ABM=∠BAM=36°

∴AM=BM

∠BEM=∠BME=72°

∴BE=BM