Question

几何题直角三角形ABC,角C是直角,BC上一点P,是BP等于AC,AC上一点Q,AQ等于CP,连

Answer 1

证明：设∠PAC=a,∠QBC=b.AQ=x,QC=y,则PB=x+y,PC=x.

∠APB=a+90°,角C是直角,则tana=x/(x+y)

tanb=y/(2x+y)

tan(a+b)=[x/(x+y)+y/(2x+y)]/[1-x/(x+y)*y/(2x+y)]

=(2x²+2xy+y²）/(2x²+2xy+y²）=1

又知0