Solve this:
Q.19. ABCD is a parallelogram, bisectors of angles A and B meet at E which lies on DC. Prove that AB = 2 AD.

Answer : 

ABCD is a parallelogram , SO we know 

AB  =  CD  and BC  =  DA  
And
AB  | | CD   and BC | | DA 


Here ∠ DAE   =  ∠ EAB ( As given AE is angle bisector of ∠ BAD ) 
And
∠ CBE  =  ∠ EBA    (  As given BE is angle bisector of ∠ ABC )

Now we take AE as transversal line and we know AB | | CD 
So 
∠ EAB  =  ∠ DEA   (  As they are alternate interior angles ) 
So,
 ∠ DEA   =  ∠ DAE   ( As we know ∠ DAE  = ∠ EAB ) 
So,
AD  = DE( As we know base angle theorem , if base angles of any triangle is equal than opposite sides are also equal ) 

And
Now we take AE as transversal line and we know AB | | CD 
So 
∠ EBA  =  ∠ CEB (  As they are alternate interior angles ) 
So,
 ∠ CBE   =  ∠ CEB  ( As we know ∠ CBE  = ∠ EBA ) 
So,
BC  = EC( As we know base angle theorem , if base angles of any triangle is equal than opposite sides are also equal ) 
Hence
AD  =  DE  =  BC  =  EC   ---- ( 1 )
We know 
AB  =  CD 
And
AB  =  DE  +  EC   ( As CD  =  DE  +  EC )

AB  =  AD  +  AD ( from equation 1  )
=> AB = 2AD ... This required proof.

Hope it helps!

Regards!