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In parallelogram ABCD two points P and Q...

In parallelogram ABCD two points P and Q are taken on diagonal BD such that DP = BQ (set figure). Show that:
`{:((i)DeltaAPD~=CQB,(ii)AP=CQ),((iii)DeltaAQB~=DeltaCPD,(iv)AQ=CP),((v)APCQ "is a parallelogram.",):}` ltbegt

Text Solution

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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")`
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