如图,AB/AD=BC/DE=AC/AE.若已知AB=bBD=3,AC=4,求CE的长

设AD=kAB=3k 则AE=kAC=4k
由bBD=3得BD=3/b=3/4
△ACE中,由余弦定理有
CE²=AC²+AE²-2AC*AE*cos∠CAE=16+16k²-32kcos∠CAE (1)
易知△ABC∽ △ADE
∠BAC=∠DAE ∴∠BAD=∠CAE
△BAD中 BD²=AB²+AD²-2AB*ADcos∠BAD=9+9k²-18kcos∠CAE
∴ 9k²-18kcos∠CAE+9-9/16=0
k²-2kcos∠CAE=-15/16
带入(1)有 CE²=16-16*15/16=1
∴ CE=1