在梯形ABCD中,AD‖EF‖BC,梯形AEFD的面积与梯形EBCF的面积相等,求证:AD^2+BC^2=2EF^2

设梯形AEFD的高为h1,梯形EBCF的高为h2,则梯形ABCD的高为(h1+h2)
梯形AEFD的面积=(1/2)(AD+EF)*h1
梯形EBCF的面积=(1/2)(BC+EF)*h2
梯形ABCD的面积=(1/2)(AD+BC)*(h1+h2)
已知梯形AEFD的面积=梯形EBCF的面积
所以,(1/2)(AD+EF)*h1=(1/2)(BC+EF)*h2
===> h2/h1=(AD+EF)/(BC+EF)…………………………………………………………（1）
那么,梯形AEFD的面积就等于梯形ABCD面积的一半
所以：(1/2)(AD+EF)*h1=(1/2)*[(1/2)(AD+BC)*(h1+h2)]
===> 2(AD+EF)*h1=(AD+BC)*(h1+h2)
===> 2(AD+EF)=(AD+BC)*[1+(h2/h1)]
===> 2(AD+EF)=(AD+BC)*[1+(AD+EF)/(BC+EF)]
===> 2(AD+EF)=(AD+BC)*(AD+BC+2EF)/(BC+EF)
===> 2(AD+EF)(BC+EF)=(AD+BC)(AD+BC+2EF)
===> 2AD*BC+2(AD+BC)EF+2EF^2=(AD+BC)^2+2(AD+BC)EF
===> 2AD*BC+2EF^2=(AD+BC)^2
===> 2AD*BC+2EF^2=AD^2+2AD*BC+BC^2
===> 2EF^2=AD^2+BC^2.