text: # 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.
  
  nhgnghnytu
  • 0
  
  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 )
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