Question

在等腰梯形ABCD中,AD//BC,对角线AC⊥AB,BD⊥CD,∠BAD=120°,若BC=8cm,求中位线EF的长

Answer 1

AC⊥AB,BD⊥CD,∠BAD=120°

∠DAC=∠ADB=30°

又∠DAC=∠ACB

∠ACB=30°

在Rt△ABC中AB=BC/2=4cm

又∠ABC=180°-120°=60°

∠DBC=30°

∠ABD=∠ABC-∠DBC=30°

∠ADB=∠ABC=30°

AD=AB=4cm

EF=(AD+BC)/2=6cm