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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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`angleBAP=angleDAP=1/2angleA" "…(1)`
Since, ABCD is a parallelogram, we have
`" "angleDAB+angleABC=180^(@)" "..(2)`
`" "(because"sum of co-interior angles on the same side of transversal is" 180^(@))`

In `DeltaABP` we have
`angleBAP+angleB+angleAPB=180^(@)" "("angle sum property")`
`implies1/2angleBAD+180^(@)-angleA+angleAPB=180^(@)`
`implies" "angleAPB-1/2angleA" "["using (1) and (2)"]`
From (1) and (3), we get `" "...(3)`
`" "angleBAP=angleAPB`
`" "BP = AB (therefore "side opposite to equal angles are equal")...(4)`
Since, opposite sides of a parallelogram are equal, we have
`" "AD=BCimplies1/2AD=1/2BC`
`implies" "1/2AD=BP" "(becauseP "is the mid-point of BC")`
`implies" "1/2AD=AB" "[because"from"(4)]`
Since, opposite sides of a parallelogram are equal, we have
`1/2AD=CDimpliesAD=2CD`
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