在平行四边形ABCD中,AC=根号2AB,求证 角COB=角DAB

问题描述:

在平行四边形ABCD中,AC=根号2AB,求证 角COB=角DAB

OC:AC = 1:2∴ OC:AB = BC:AC = 1:√2∴ △ACB ∽ △ABO∴ ∠ACB = ∠ABO∴ ∠COB = ∠CAB + ∠ABO (△AOB,外角定理) = ∠CAB + ∠ACB (上面证明的,∠ACB = ∠ABO) = ∠CAB + ∠CAD (AD‖BC,∴...