1. ## * Geometry Help. Two questions. ASAP help please.

Write proofs in two-column form. (One side is the Statements and the other is the Reasons why.)

1a) The diagram shows three squares and an equilateral triangle.
Prove: AE=FC=ND

Diagram:

1b) Use the results in problem 1a to prove that Triangle FAN is equilateral.

Please explain to me in details..I'm going to have a test soon, and I do not understand this! All help are greatly appreciated! Thanks!

3. Originally Posted by MagixAries

Let's try @Aborted @Lehsyrus @666HiddenMaster666 @arunforce @Dave84311

4. I don't think I've ever done a two column proof, but the three squares are congruent and the triangle is equilateral, and each square shares an edge with the triangle so they simply have to be the same length.

I guess a two column proof would go like this

DC = CE = ED, because the triangle is equilateral

AB = BC = CD = DA = CN = NM = ME = EC = EF = FG = GD = DE, because the squares share a side with a triangle, and by definition a square had equal length sides

therefore sqrt(AB^2+AB^2) = AE & sqrt(AB^2 + AB^2) = CF & sqrt(AB^2 + AB^2) = DN, because triangle and each segment is equal to any other segment.

AE = CF = DN, because they are all computed using the same formula

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6. Originally Posted by arunforce
I don't think I've ever done a two column proof, but the three squares are congruent and the triangle is equilateral, and each square shares an edge with the triangle so they simply have to be the same length.

I guess a two column proof would go like this

DC = CE = ED, because the triangle is equilateral

AB = BC = CD = DA = CN = NM = ME = EC = EF = FG = GD = DE, because the squares share a side with a triangle, and by definition a square had equal length sides

therefore sqrt(AB^2+AB^2) = AE & sqrt(AB^2 + AB^2) = CF & sqrt(AB^2 + AB^2) = DN, because triangle and each segment is equal to any other segment.

AE = CF = DN, because they are all computed using the same formula
For #23,I got this:
1. AD=DE=GE=EC=NC=DC; def of regular polygon & def. of squares
2. DE=DE, EC=EC, DC=DC; reflex. prop.
3. angle CDE, ECD, and DEC are all 60 degree; def. of equilateral triangle
4. triangle FEC is congruent to triangle ADE is congruent to triangle DCN; SAS postulate
5. AE=FC=ND; CPCTC
and then I got stuck on how to prove triangle FAN as a equilateral triangle :/

7. Originally Posted by MagixAries

For #23,I got this:
1. AD=DE=GE=EC=NC=DC; def of regular polygon & def. of squares
2. DE=DE, EC=EC, DC=DC; reflex. prop.
3. angle CDE, ECD, and DEC are all 60 degree; def. of equilateral triangle
4. triangle FEC is congruent to triangle ADE is congruent to triangle DCN; SAS postulate
5. AE=FC=ND; CPCTC
and then I got stuck on how to prove triangle FAN as a equilateral triangle :/
It is equilateral because the squares are all touching and all the squares are equal

8. Originally Posted by MagixAries