SOLUTION: In a parallelogram ABCD,the bisector of angle A also bisects BC at X.Prove that AD = 2AB

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Question 129130: In a parallelogram ABCD,the bisector of angle A also bisects BC at X.Prove that AD = 2AB
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
In a parallelogram ABCD,the bisector of angle A also bisects BC at X.Prove that AD = 2AB

Given:  parallogram ABCD, angle BAX = angle DAX, BX = CX

To prove: AD = 2AB



Extend AX and DC till they meet at Y 



1. Angle BAX = Angle CYX
2. Angle BAX = Angle DAY
3. Angle CYX = Angle DAY
4. Triangle DAY is isosceles
5. Angle B = Angle D
6. Triangle ABX similar to Triangle DAY
7. BC = 2BX
8. AY = 2AX
9. DY = 2AB
10. AD = DY
11. AD = 2AB

1. Alternate interior angles when transversal AY 
   cuts parallel lines AB and DY
2. Given that AX bisects angle A 
3. Things equal to the same thing are equal to each other
4. Base angles equal
5. Opposite interior angles of a parallelogram
6. Two angles equal in each
7. Given that AX bisects BC
8. Two intersecting transversals BC and AY between 
   two parallel lines AB and DY are divided in the same ratio.
9. Corresponding parts of similar triangles are in the same ratio.
10. Legs of isosceles triange DAY are equal 
11. Things equal to the same thing are equal to each other.
  
Note: Incidentally, you could also prove that 
Angle A = Angle C = 60, Angle B = Angle D = 120
but that was not asked for. 

Edwin


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