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In Figure, P is the mid-point of side BC of a parallelogram ABCD such that \angle BAP=\angle DAP. Prove that AD = 2CD

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Solution.
Given : ABCD is a parallelogram, P is a mid-point of BC such that \angle BAP = \angle DAP.
To prove : AD = 2CD
Proof : Here ABCD is a parallelogram

\therefore AD\parallel BC and AB is transversal
\angle A+\angle B=180^{\circ}               {sum of co-interior angles}

\angle B=180-\angle A                
…..(1)
In
\triangle ABP,  
\angle PAB+\angle BPA+\angle B=180^{\circ}      {using angle sum property of triangle}
Putting the values,
\frac{1}{2}\angle A+\angle BPA+(180^{\circ}-\angle A)=180^{\circ}
\angle BPA-\frac{\angle A}{2}=0
\angle BPA=\frac{\angle A}{2}                      …..(2)
\Rightarrow \angle BPA=\angle BAP

AB = BP           {opposite sides of equal triangle}
2AB = 2BP            {multiply both sides by 2}
2AB = BC             { P is midpoint of BC}
2CD = AD
      { ABCD is parallelogram   AB = CD and BC = AD}                       

Hence proved


 

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