If xa=yb=zc, then xy+yz+zx=?
xa=yb=zc=k x=ak y=bk z=ck (a+b+c)2=a2+b2+c2+2(ab+bc+ca) xy+yz+xz=(ab+bc+ca)k2 =x2a2(ab+bc+ca) =x22a2[(a+b+c)2−(a2+b2+c2)] =x22a2[(a+b+c)2−(x2+y2+z2k2)] =x2(a+b+c)2−a2(x2+y2+z2)2a2