# Similarity between triangles

1. Jan 19, 2008

### disregardthat

Hi, do anyone know a proof of this converse:

"If, A,B,C,D,E and F are points in the plane and $$\frac{AB}{BC}=\frac{DE}{EF}$$, then triangles ABC and DEF are similar."

Last edited: Jan 19, 2008
2. Jan 19, 2008

### EnumaElish

What is your definition of similar?

3. Jan 19, 2008

### disregardthat

Oh, with similiar I mean that the triangles are equiangular.

With $$\angle ABC = \angle DEF$$ and $$\angle ABC = \angle DEF$$

Last edited: Jan 19, 2008
4. Jan 19, 2008

### EnumaElish

Do you mean /_ABC = /_ DEF and /_BCA = /_EFD ? (You repeated yourself.)

Last edited: Jan 19, 2008
5. Jan 19, 2008

### disregardthat

Yes, I meant the last thing you said. (Getting late.)

6. Jan 19, 2008

### jdg812

No one knows a proof, because such a proof doesn’t exist.

PS
But in the case AB:BC:CA = DE:EF:FD, proof is trivial

7. Jan 19, 2008

### disregardthat

Forgot to mention that they share the angle between the sides AB and BC, and DE and EF.

Oh, and the expression should not look like that either...

It's $$\frac{AB}{DE}=\frac{EF}{BC}$$

EDIT: Anyway, I have proven it now...

Last edited: Jan 19, 2008
8. Jan 19, 2008

### jdg812

Still not enough for similarity... :rofl:

If you forget as well to mention that equal angles are /_ABC and /_ DEF, proof is trivial.
If other angles, proof doesn't exist.

9. Jan 19, 2008

### disregardthat

There has been to many faults here. I will sum it up and see what I find out!

10. Jan 19, 2008

### jdg812

After your editions (bolded) proof is not possible

It must be

$$\frac{AB}{DE}=\frac{BC}{EF}$$

OR

$$\frac{AB}{BC}=\frac{DE}{EF}$$

Last edited: Jan 19, 2008
11. Jan 19, 2008

### disregardthat

Ok, I have summed it up now. I am trying to prove the converse of the intersecting chords theorem, in the case where the point of intersection is inside the circle...

We have that the line segments AB and CD meet at X. So the opposite angles at X are equal. We translate the triangles to the triangle with sides XAD, XCB and with AB and DC joined. Now the angle at X is a, and $\frac{XA}{XC}=\frac{XD}{XB}$. This is a sufficient condition for the triangles XAC and XDB to be similar, as corresponding sides have the same ratio, and the included angle is equal. Thus is the angle XDB equal to the angle XAC, and by the converse of the angles subtended by the same arc theorem, ACBD are concyclic points, so AB and CD are chords in a circle. Does this look ok to you? The other cases of the converse goes something in the same way.

Last edited: Jan 19, 2008