ice_kite
幼苗
共回答了17个问题采纳率:94.1% 举报
不妨设,正方形边长1,角MCB=x,
作:PQ垂直CD于Q,连接DN
MB=tanx,所以:BN=MB=tanx
PB=sinx,
CP=cosx
PQ=CP*cosx=(cosx)^2
QC=CP*sinx=sinx*cosx
DQ=1-QC=1-sinx*cosx
NC=BC-BN=1-tanx
角CBP=(pi/2)-x
DN^2=DC^2+CN^2=1+(1-tanx)^2=(tanx)^2-2tanx+2
DP^2=DQ^2+PQ^2=(cosx)^4+(1-sinx*cosx)^2
PN^2=PB^2+BN^2-2PB*BN*cos角CBP
=(sinx)^2+(tanx)^2-2sinx*tanx*sinx
=(sinx)^2+(tanx)^2-2(sinx)^2*tanx
DP^2+PN^2=(cosx)^4+(1-sinx*cosx)^2+(sinx)^2+(tanx)^2-2(sinx)^2*tanx
=(cosx)^4+(sinx)^2*(cosx)^2-2sinxcosx+1+(sinx)^2+(tanx)^2-2(sinx)^2*tanx
=(cosx)^2((cosx)^2+(sinx)^2)+(sinx)^2-2sinxcosx+1+(tanx)^2-2(1-(cosx)^2)*tanx
=(cosx)^2+(sinx)^2-2sinxcosx+1+(tanx)^2-2tanx+2sinxcosx
=2+(tanx)^2-2tanx
=DN^2
所以:三角形DPN为直角三角形
PD垂直PN
1年前
9