Question

在正方形ABCD中,点M为AD上一点,BN平分角CBM,交CD于点N,求证BM=CN+AM

Answer 1

∠CBN=∠MBN=1/2*∠CBMAM=AB*tan∠ABM=AB/tan(2∠CBN)BM=AB/cos(∠ABM)=AB/sin(2∠CBN)CN=BC*tan∠CBN简单写∠CBN=θAM+CN=AB*[tanθ+1/tan(2θ)]=AB*[2(sinθ)^2+cos(2θ)]/sin(2θ)cos(2θ)=(cosθ)^2+(sinθ)^2AM...