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## #1 2007-11-08 09:23:32

tony123
Member
Registered: 2007-08-03
Posts: 196

### hexagon

Each interior angle of the hexagon in the figure is 120°. Prove that

AB + FA = CD + DE.

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## #2 2007-11-09 23:24:40

gyanshrestha
Member
Registered: 2007-11-06
Posts: 41

### Re: hexagon

if it is regular hexagon is posssible to prove. in that case all sides are equal and proof will easily follow.
if not the case then i think it is not possible.

sorry for inconvinience

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## #3 2007-11-09 23:32:34

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: hexagon

It does not have to be a regular hexagon. All Ive managed so far is to show that opposite sides must be parallel.

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## #4 2007-11-10 19:11:36

gyanshrestha
Member
Registered: 2007-11-06
Posts: 41

### Re: hexagon

is it possible that opposite sides are parallel and their sum are equal?
then it must be BF=CE, then is it possible?

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## #5 2007-11-11 00:08:48

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: hexagon

No. Opposite sides do not have to be equal. The question is, prove that the sum of two adjacent sides is equal to the sum of the opposide two adjacent sides.

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## #6 2007-11-11 00:36:31

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: hexagon

This is another example of a hexagon with all interior angles equal to 120° but with sides of different lengths. Its been drawn to scale (hopefully) so you can print the image and measure its sides directly. (Presumably Tonys hexagon is also drawn to scale so you could also do the same thing with his polygon.)

Indeed, the result sum of adjacent sides = sum of opposite adjacent sides is true for all even-sided polygons (not just the hexagon) whose interior angles are all equal.

Last edited by JaneFairfax (2007-11-11 00:41:21)

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## #7 2007-11-16 09:28:41

tony123
Member
Registered: 2007-08-03
Posts: 196

### Re: hexagon

equilateral triangles at two opposite ends

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## #8 2007-11-16 09:37:33

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

### Re: hexagon

tony123 wrote:

equilateral triangles at two opposite ends

Now that youve pointed it out

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