# Thread: A proof is needed

1. ## A proof is needed

ABCD is a square.
BEC and DCF are equilateral triangles on the east and south side of the square respectively.

Having proved ECD=BCF I now need to prove that ED=EG. Got as far as FG=BE and angle CFE= angle EBC

2. Hello, Mukilab!

$\displaystyle ABCD$ is a square.

$\displaystyle BEC$ and $\displaystyle DCF$ are equilateral triangles on the east and south side of the square, resp.

Having proved $\displaystyle \angle ECD\,=\, \angle BCF$, I now need to prove that $\displaystyle ED=EG.$ .?
Code:
     A             B
o - - - - - o
|           |  *
|           |     *
|           |        o E
|           |     *
|           |  *
o - - - - - o
D *         * C
*       *
*     *
*   *
* *
o
F

Where is point $\displaystyle G$ ?

3. I don't know exactly but if BFGE is a parallelogram then I guessed it would be somewhere to the south-east of F

(Doesn't show it in any sort of diagram)

4. Originally Posted by Mukilab
I don't know exactly but if BFGE is a parallelogram then I guessed it would be somewhere to the south-east of F

(Doesn't show it in any sort of diagram)
If BFGE is a parallelogram, then BF=GE,
so you only need to show that ED=BF as that means ED=GE.