# A proof is needed

• Jun 5th 2010, 11:51 AM
Mukilab
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
• Jun 5th 2010, 02:25 PM
Soroban
Hello, Mukilab!

Quote:

\$\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\$ ?

• Jun 6th 2010, 12:41 AM
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)
• Jun 6th 2010, 02:39 AM
Quote:

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.