∵AD∥BC,
∴△MAD∽△MBC,
∴
AD |
BC |
MA |
MB |
1 |
3 |
∴MB=3MA.设MA=2x,则MB=6x.
∴AB=4x.
∵BE=3AE,
∴BE=3x,AE=x.
∴BE=EM=3x,E为MB的中点.
又∵CE⊥AB,
∴CB=MC.
又∵MB=MC,
∴△MBC为等边三角形.
∴∠B=60°;
(2)延长BA、CD相交于点F,如图2:
∵AD∥BC,
∴△FAD∽△FBC,
∴
S△FAD |
S△FBC |
AD |
BC |
1 |
9 |
设S△FAD=S3=a,则S△FBC=9a,S1+S2=8a,
又∵2S1=3S2,
∴S1=
24 |
5 |
16 |
5 |
∵△EFC与△CEB等高,
∴
FE |
EB |
S△FEC |
S△ECB |
S3+S2 |
S1 |
7 |
8 |
设FE=7k,则BE=8k,FB=15k,
∴FA=
1 |
3 |
∴AE=7k-5k=2k.
∴
BE |
AE |