1.S∆ADE相似于∆ABC,设∆ABC高H,S∆ADE高h,DE=a,BC=b,且可得a:b=H:h,得H=hb/a
又0.5ah=4,0.5b(H-h)=24
所以条件连立得b=3a,亦可得AD:DB=1:3
因为S∆BDE与S∆ADE等高,所以S∆BDE:S∆ADE=BD:AD=3:1=S∆BDE:4,得S∆BDE=12
2.如果abc三者间比值必须是固定的,那么DE不一定平行与BC,不妨假设DE平行与BC,看其比值是否固定即可推断
S∆ADE:S∆BDE=a:b得AB:BD=a:b
S∆BDE:∆BCE=b:c得DE:BC=b:c
又AD:AB=DE:BC得a/(a+b)=b/c