Question

hope you answer my question

Answer 1

ΔPBQ & ΔPCD  are similar, as the sides are parallel...
As BP : PC = 1 : 2,      Area(ΔPCD) = 2² * Area(ΔPBQ)

Draw a perpendicular to ABQ passing through P meeting AB a P' & CD at P''.  P'P'' is the altitude of the parallelogram. It is also the altitude of ΔCBQ.  PP' is the altitude of ΔPBQ.

Again from similar triangles principles, we infer that  P'P'' = 3 * P'P  (or, P"P=2*P'P).

Area(ΔPBQ) = 1/2 BQ * PP'
Area(ΔCBQ) = 1/2 BQ * P''P' = 3 * Area(ΔPBQ)
                     = Area(ΔPBQ) + Area(ΔCPD)
=>  Area(ΔPBQ) = 1/2 * 20 = 10 cm²
=> Area (ΔCPD) = 40 cm²

From similar triangles(PBQ,CPD), we get    BQ = CD/2  = AB/2

Area(ABCD) = AB * altitude = AB * P''P' = 2 * Area(ΔCBQ) = 2*30 cm²
       = 60 cm²

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Q5

ΔOAB :   as ∠AOD = ∠BOD :   ratio OA/OB = AD / BD     ---(1)
ΔOBC:    as ∠BOE = ∠COE :   ratio OB /OC = BE/CE     -- (2)
ΔOCA:   as  ∠COF = ∠AOF,    ratio  OC/OA = CF/AF    --- (3)

Multiply LHS of (1), (2), & (3):     1 = AD * BE * CF/ (BD * CE * AF)

That is the proof needed.