做D做D'D垂直于AB于D'点,设对角线相交于M点
设中卫线为L=（AB+CD)/2,高为H
又因对角线AC⊥BD
显然△DBD'∽△ABM∽△DMC
则BM/AB=DM/DC=BD'/BD
=>BM/AB=DM/DC=BD'/BD=(BM+DM)/(AB+DC)=BD/2L
则BD'=BD²/2L=72/L
由于△S=LH
另外△S=S△ABD+S△DCB=BD*AM/2+BD*CM/2=AC*BD/2=30
又H²+BD'²=BD²=144
H²+(72/L)²=144
LH=30
消去H可得
(30/L)²+(72/L)²=144
得L=13/2