ABCD-trapez oarecare
fie AM_|_DC,
AM-h
fie BN_|_DC
DC-h
AM=DC=2√3
MN=AB=6
ΔAMD, mas<M=90
mas<ADC=45 ⇒mas<DAM=45 ⇔ΔAMD dreptunghic isoscel DM=AM=2√3
AD²=DM²+AM²= (2√3)²+ (2√3)²= 12+12=24 ⇒AD=2√6
ΔBNC, mas<N=90
mas<BCD=30
sin<BCD=30=1/2
sin<BCD=BN/BC= 2√3/BC
2√3/BC=1/2
BC=2√3*2/1
BC=4√5
