# proof

• May 9th 2012, 06:43 AM
Mhmh96
proof
If ABCDEF a regular hexagon,and FCGH is a square and AB=4 , FC=8 prove that the triangle IED is similar to the triangle IGH .

http://store2.up-00.com/May12/8pg74506.jpg
• May 9th 2012, 06:56 AM
emakarov
Re: proof
This reminds me a joke. A visitor to Odessa, a "capital of Soviet humor," asks a local resident: "If I go down this street, will there be a railway station?" The local replies, "The station will be there even if you don't go." In this case, triangles IED and IGH are similar even if AB is not 4.

It is easy to show that the two triangles have their corresponding angles equal since ED is parallel to HG.
• May 10th 2012, 07:46 AM
Mhmh96
Re: proof
If we prove that all angles are equal ,is that enough to prove that these two triangles are similar ?
• May 10th 2012, 07:49 AM
Sylvia104
Re: Proof
Quote:

Originally Posted by Mhmh96
If we prove that all angles are equal ,is that enough to prove that these two triangles are similar ?

Yes.
• May 10th 2012, 08:00 AM
Mhmh96
Re: proof
But why they are corresponding angles???,they looks alternate angles
• May 10th 2012, 08:33 AM
emakarov
Re: proof
Quote:

Originally Posted by emakarov
It is easy to show that the two triangles have their corresponding angles equal since ED is parallel to HG.

Quote:

Originally Posted by Mhmh96
But why they are corresponding angles???,they looks alternate angles

I used the word "corresponding" in its general meaning to mean "matching." E.g., angle IED corresponds to angle IGH and not angle HIG. I agree that IED and IGH are alternate interior angles, not corresponding in the technical sense of angles produced by a transversal.