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 BE=BM

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
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   BE=BM

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