given
(pa+qb):(pc+qd)::(pa−qb):(pc−qd)
⇒pa+qbpc+qd=pa−qbpc−qd
pa+qbpa−qb=pc+qdpc−qd
now using compodendo and dividendo rule [a/b=c/d implies (a+b)/(a-b)=(c+d)/(c-d)] we get
(pa+qb)+(pa−qb)(pa+qb)−(pa−qb)=(pc+qd)+(pc−qd)(pc+qd)−(pc−qd)
2pa2qb=2pc2qd⇒ab=cd
therefore a:b::c:d