ABCD is a parallelogram and P and Q are lie on BD such that
DP = BQ
(i) We have to show,
`" "DeltaAPD~= DeltaCQB`
Now, in `DeltaAPDand DeltaCQB,` we have
`because{:{(DP=AQ,("give")),(AD=BC,("opposite sides of parallelogram")),(angle1=angle2,("alterate angle")):}`
`therefore" "DeltaPAD~=DeltaCQB" "("by SAS")`
(ii) `"Since"" "DeltaAPD~=DeltaCQB" "("just proved")`
`therefore" "AP=CQ`
(iii) Here, we have to show, `DeltaAQB~=DeltaCPD`
Now, in `DeltaAQBand DeltaCPD,` we have
`because{:{(BQ=DP,("given")),(AB=CD,("opposite sides of parallelogram")),(angle3=angle4,("alternate angles")):}`
`therefore" "DeltaAQB~=DeltaCPD" "("by SAS")`
(iv) `"since"" "DeltaAQB~=DeltaCPD`
`therefore" "AQ=Cp`
(v) Now, since AP = CQ
and AQ = CP
`thereforesquareAPCQ "is a parallelogram." " "(because"pairs of opposite sides are equal")`