Question

Please can someone solve this for me

Answer 1

in ∆abp and ∆cpd,
< apb = < cpd ( vertically opposite angles )
< bap = < dcp ( alternate interior angles because ab || cd )
by aa similarity ∆ apb ~ ∆ cpd
so
A (∆ apb)/ A (∆ cpd) = (ap/cp)^2 = (pb/pd)^2 = (ab/cd)^2
[ the ratios of areas of two similar triangles is equal to the ratio of square of its corresponding sides.]
hence ,proved...