Question

如图,在正方形ABCD中,Q在CD上,且DQ=QC,P在BC上,且AP=CD+CP.求证：AQ平分∠DAP

Answer 1

证：设正方形边长为a,

AP=CD+CP=a+CP,AP^2=a^2+2aCP+CP^2,

AP^2=a^2+(a-CP)^2=2a^2-2aCP+CP^2,

CP=a/4,AP=5a/4；

PQ^2=AP^2+AQ^-2AP*AQ*cos∠PAQ

cos∠PAQ=[25a^2/16+(a^2+a^2/4)-(a^2/4+a^2/16)]/2(5a/4*a√5/2)

=2√5/5

cos∠PAD=a/(a√5/2)=2√5/5

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