Question

梯形ABCD中AB平行于CD,角ADC+角BCD=90度,以AD,AB,BC为斜边向形外作等腰直角三角形,其面积分别是S

Answer 1

CD= 3AB

作BE//AD 交DC于E ,因,∠ADC+∠BCD=90°,故∠EBC=90°,

S1=AD^2/2 ,S3=BC^2/2 , S2=AB^2/2 , S1+S3=4S2 ,

即 AD^2+BC^2 = 4AB^2

又 CD =AB+√AD^2+BC^2 =3 AB