如题所述
过作AM∥EF,交BC于M,BN∥GH,交CD于N,
由平行四边形AMFE和BNHG得AM=EF,BN=GH,
且AM⊥BN,
∴∠BAM+∠AMB=∠CBN+∠AMB,
∴∠BAM=∠CBN,
又∵∠ABC=∠C=90°,
∴△ABM∽△BCN,
∴AM/BN=AB/BC=a/b
∴EF/GH=a/b
解:作CM∥EF,BK∥GH,则CM=EF,BK=GH,
且CM⊥BK。∵∠MCD=∠KBC,∴Rt△MCD∽Rt△KBC,
CM/BK=CD/BC=a/b, ∴EF/GH=a/b.