Question 129130
In a parallelogram ABCD,the bisector of angle A also bisects BC at X.Prove that AD = 2AB
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Given:  parallogram ABCD, angle BAX = angle DAX, BX = CX

To prove: AD = 2AB

{{{drawing(400,375,-1,6,-1,6,
line(0,0,2,0), line(2,0,4.618802154,4), line(4.618802154,4,2.618802154,4),
line(2.618802154,4,0,0),line(0,0,3.3,2),
locate(-.1,-.1,A), locate(2,-.1,B), locate(4.6,4.3,C),locate(2.5,4.3,D) 
locate(3.3,2,X)
 )}}}

Extend AX and DC till they meet at Y 

{{{drawing(400,375,-1,8,-1,8,
line(0,0,2,0), line(2,0,4.618802154,4), line(6.618802154,4,2.618802154,4),
line(2.618802154,4,0,0),line(0,0,6.618802154,4),
locate(-.1,-.1,A), locate(2,-.1,B), locate(4.6,4.4,C),locate(2.5,4.4,D) 
locate(3.3,2,X),locate(6.7,4.4,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</pre>