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In figure, P is the mid-point of side BC...

In figure, P is the mid-point of side BC of a parallelogram ABCD such that `angleBAP=angleDAP`. Prove that AD = 2CD.

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Given In a parallelogram ABCD, P is mid-point of BC such that `angleBAP=angleDAP`.
To prove `" "AD=2CD" "`
Proof Since, ABCD is a parallelogram. ltBrgt So, AD`||`BC and AB is transversal, then
`" "angleA+angleB=180^(@)" "`[sum of cointerior angles is `180^(@)`]
`rArr" "angleB=180^(@)-angleA" "...(i)`
In `Delta`ABP, `" "anglePAB+angleB+angleBPA=180^(@)" "`[by angle sum property of triangle]
`rArr" "(1)/(2)angleA+180^(@)-angleA+angleBPA=180^(@)" "` [from Eq. (i)]
`rArr" "angleBPA-(angleA)/(2)=0`
`rArr" "angleBPA=(angleA)/(2)" "...(ii)`
`rArr" "angleBPA=angleBAP`
`rArr" "AB=BP` [opposite sides of equal angles are equal]
On multiplying both sides by 2, we get
`" "2AB=2BP`
`rArr" "2AB=BC" "` [since P is the mid-point of BC] ltBrgt `rArr" "2CD=AD`
`" "`[since, ABCD is a parallelogram, then AB=CD and BC =AD]
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